Test 4 Review Material

Below are resources to help you practice and prepare for the test 4 in this course.

Test 3 will cover the material covered in Chapter 6 of this textbook:

understanding the process of optimizaiton
findind opimal solutions in constrained problems
identifying feasible regions for solutions given question context

The test will include:

5 short answer (10pts)
3 optimization problems (40pts)

Test Instructions

Below are the general instructions for all tests

You will get a chance to retake this. Your highest score counts toward your final grade.
Follow the guidance for each part and show all work for full credit.
Non-graphical calculators are allowed.
One page (two sides) of handwritten (or font size 8) notes are allowed.
The equation sheet from the textbook is included as part of the test pack (it doesn’t count as note pages)

Short-Answer Optimization Questions

What is the “objective function” in an optimization problem?

Solution

It is the function you want to maximize or minimize—e.g., growth, cost, carbon storage, habitat value.

Why is defining the objective function often the hardest step in real environmental problems?

Solution

Because you must decide exactly what you are optimizing and how competing processes (benefits vs. penalties, diminishing returns, constraints) interact mathematically.

How do you find critical points of a differentiable function?

Solution

Differentiate the objective function and set the derivative equal to zero; solve for the variable(s).

What does the second derivative tell you about a critical point?

Solution
If $f''(x)>0$: local minimum.
If $f''(x)<0$: local maximum.
If $f''(x)=0$: inconclusive.

What should you do if the derivative never equals zero on the domain?

Solution

There are no interior critical points; the optimum must occur at a boundary or the function is monotonic.

What if the derivative equals zero only at a point outside the feasible domain?

Solution

The critical point is thrown out. The optimum must lie at a domain boundary or another feasible critical point.

What should you always check when a problem includes a constraint?

Solution

Check whether the unconstrained optimum is feasible. If not, the optimum lies on the boundary.

Why do constrained optimization problems often require evaluating endpoints?

Solution

Because constraints restrict the domain; global optima often occur at boundaries, not at interior points.

Why must you convert multivariable problems (with a constraint) into one variable before differentiating?

Solution

Because the variables are not independent; substituting the constraint ensures the derivative respects the relationship between variables.

What does it mean if a critical point yields a negative value for a physical quantity (e.g., width, depth)?

Solution
It is physically impossible and must be rejected.

Why can a function have a local maximum that is not the global maximum?

Solution

Local maxima only beat nearby values; global maxima require comparing all feasible values, including boundaries.

Give one reason why the optimum might occur at the minimum allowed value of the variable.

Solution

If the objective decreases as the variable increases, the smallest allowed value gives the best outcome.

Give one reason why the optimum might occur at the maximum allowed value.

Solution

If the objective increases with the variable and there is no interior critical point, the endpoint yields the maximum.

A function is always increasing over the domain. Where is the maximum?

Solution

At the right endpoint.

A function is always decreasing over the domain. Where is the minimum?

Solution

At the right endpoint (and maximum at the left endpoint).

Why does concavity simplify optimization?

Solution

If concave down everywhere → any critical point is a global maximum.
If concave up everywhere → any critical point is a global minimum.

What should you do if the second derivative at a critical point is zero?

Solution
Use a first derivative sign chart, higher derivatives, or direct function comparison. The point may be max, min, or neither.

Why must physical context be checked even after math gives a critical point?

Solution
Because the mathematically correct point may be unrealistic (negative depth, unsafe height, impossible density, etc.).

How can you tell if the objective function incorrectly models the system?

Solution
If the model predicts impossible or nonsensical optimum values (infinite spacing, negative biomass), the formulation is wrong.

What does it mean if two designs give nearly identical objective values?

Solution
The exact optimum may not matter; choose based on cost, practicality, or robustness. Sensitivity trumps precision.

Why can a local maximum be ecologically meaningless?

Solution
If a larger feasible value exists elsewhere or if the local maximum lies in a biologically irrelevant part of the domain.

In a constrained pond-design problem, why is the shallowest allowed depth sometimes optimal?

Solution
Because some models reward surface area more than depth, so minimum depth maximizes habitat or productivity.

Why might an interior critical point be a minimum rather than a maximum?

Solution
Because the second derivative is positive (concave up). The global maximum might lie at a boundary.

Why should you verify your derivative before solving the optimization problem?

Solution
A single differentiation error leads to entirely incorrect critical points and interpretations.

What are the three steps for finding a global maximum on a closed interval?

Solution

Find interior critical points.
Evaluate the objective at all critical points.
Evaluate endpoints and compare all values.

Practice: Optimization in Environmental Systems

Work through the following 10 new problems to deepen your fluency with applied optimization, constraints, and interpretation.

1. Optimizing Leaf Nitrogen for Maximum Photosynthesis

Leaf nitrogen concentration $N$ strongly influences photosynthetic rate, but only up to a point.
A simplified model:

\[ P(N) = 12N - N^2 \]

where:

$N$ = leaf nitrogen (% of dry mass)
$P(N)$ = relative photosynthetic rate

Your Task

Find the nitrogen level that maximizes photosynthesis.
Confirm with the second derivative.
Explain why both low and high nitrogen can be suboptimal.

Show Solution

Differentiate:

\[ P'(N) = 12 - 2N. \]

Set to zero:

\[ 12 - 2N = 0 \Rightarrow N = 6. \]

Second derivative:

\[ P''(N) = -2 < 0. \]

Thus $N = 6\%$ maximizes photosynthesis.

Interpretation:
Low nitrogen limits pigment formation; excess nitrogen leads to costly maintenance respiration.
The model identifies the efficient “sweet spot.”

2. Designing a Riparian Buffer Strip for Maximum Nitrate Removal

A riparian buffer is planted along a stream to reduce nutrient runoff.
Nitrate removal per meter of stream is modeled as:

\[ R(w) = 20w - 0.4w^2 \]

where $w$ = buffer width (m).

Your Task

Find width maximizing nitrate removal.
Confirm it with the second derivative.
Interpret ecologically.

Show Solution

\[ R'(w) = 20 - 0.8w. \]

Set to zero:

\[ 20 - 0.8w = 0 \Rightarrow w = 25. \]

Second derivative:

\[ R''(w) = -0.8 < 0 \Rightarrow \text{maximum.} \]

Interpretation:
Buffers too narrow miss pollutants; buffers too wide add little new plant/soil interface.
Twenty-five meters maximizes filtration performance.

3. Optimizing Grazing Intensity for Meadow Restoration

Meadow biomass under light grazing increases, but declines under heavy grazing.
A restoration team models biomass as:

\[ B(g) = 80g - 5g^2, \]

where $g$ is grazing intensity (sheep per hectare).

Your Task

Find the optimal grazing intensity.
Confirm with the second derivative.
Explain how this relates to “intermediate disturbance.”

Show Solution

\[ B'(g) = 80 - 10g. \]

Set to zero:

\[ 80 - 10g = 0 \Rightarrow g = 8. \]

Second derivative:

\[ B''(g) = -10 < 0. \]

Interpretation:
Very low grazing allows dominance by tall competitors; very high grazing suppresses all vegetation.
An intermediate level (8 animals/ha) maximizes overall biomass.

4. Sizing a Biochar Kiln to Maximize Net Carbon Benefit

A small biochar kiln processes biomass at rate:

\[ C(s) = 50s - 0.5s^2, \]

where $s$ is kiln size (m³) and $C(s)$ is net carbon benefit per batch.

Your Task

Find the optimal kiln size.
Use second derivative to confirm.
Explain why both too small and too large kilns are problematic.

Show Solution

\[ C'(s) = 50 - s. \]

Set to zero:

\[ 50 - s = 0 \Rightarrow s = 50. \]

Second derivative:

\[ C''(s) = -1 < 0. \]

Interpretation:
Small kilns underutilize heat; very large kilns lose heat to inefficiency and incomplete pyrolysis.
A 50 m³ kiln balances throughput and efficiency.

5. Solar Farm Layout: Maximize Energy per Unit Land Area

Panel spacing $d$ affects shading and reflection losses.
Energy per panel is modeled as:

\[ E(d) = 200d - 10d^2. \]

Your Task

Find spacing that maximizes energy production.
Check with second derivative.
Explain physically.

Show Solution

\[ E'(d) = 200 - 20d. \]

Set to zero:

\[ 200 - 20d = 0 \Rightarrow d = 10. \]

\[ E''(d) = -20 < 0. \]

Interpretation:
Panels too close shade each other; too far wastes land.
Ten meters gives optimal solar spacing.

6. Terraced Hillside Farming: Minimizing Soil Loss

A farmer designs terrace spacing $x$ to minimize yearly soil loss:

\[ L(x) = \frac{100}{x} + 0.5x, \]

where:

$\frac{100}{x}$: erosion reduced by closer spacing
$0.5x$: cost of constructing/maintaining terraces

Your Task

Find $x$ minimizing soil loss.
Use the second derivative to confirm.
Interpret tradeoffs.

Show Solution

\[ L'(x) = -\frac{100}{x^2} + 0.5. \]

Set to zero:

\[ 0.5 = \frac{100}{x^2} \Rightarrow x^2 = 200 \Rightarrow x = \sqrt{200} \approx 14.14. \]

Second derivative:

\[ L''(x) = \frac{200}{x^3} > 0. \]

Thus $x \approx 14.14$ m minimizes soil loss.

Interpretation:
Tight spacing reduces erosion but is expensive; wide spacing saves cost but erodes more.
The optimum balances erosion control and cost.

7. Designing a Wildlife Pond: Maximizing Habitat With a Volume Constraint

Restoration ecologists are designing a circular wildlife pond to support amphibians, birds, and aquatic plants.
Wide, shallow ponds offer the best habitat.

Let:

$r$: pond radius (m)
$h$: pond depth (m)

Volume requirement:

\[ \pi r^2 h = 200. \]

Habitat quality:

\[ H(r,h) = \pi r^2 - 4h. \]

Minimum ecological depth:

\[ h \ge 0.5. \]

Show Solution

Step 1: Express $H$ in terms of $r$

Use the volume constraint:

\[ h = \frac{200}{\pi r^2}. \]

Substitute into $H$:

\[ H(r) = \pi r^2 - 4\left(\frac{200}{\pi r^2}\right) = \pi r^2 - \frac{800}{\pi r^2}. \]

Step 2: Differentiate

Rewrite to differentiate cleanly:

\[ H(r) = \pi r^2 - \frac{800}{\pi} r^{-2}. \]

Differentiate:

\[ H'(r) = 2\pi r + \frac{1600}{\pi r^3}. \]

Step 3: Look for critical points

Set $H'(r)=0$:

\[ 2\pi r + \frac{1600}{\pi r^3} = 0. \]

Both terms are positive for all $r>0$.
Therefore, this equation has no solution.

Conclusion:
There is no interior critical point.
Habitat quality increases as radius increases.

Thus the maximum must occur on the boundary of the feasible region.

Step 4: Apply the depth constraint

Minimum allowable depth:

\[ h = 0.5. \]

Use the volume constraint to solve for $r$:

\[ \pi r^2 (0.5) = 200 \]

\[ r^2 = \frac{400}{\pi} \]

\[ r^* = \sqrt{\frac{400}{\pi}} = \frac{20}{\sqrt{\pi}} \approx 11.28\ \text{m}. \]

Final Optimal Pond Design

Optimal depth:
\[ h^* = 0.5\ \text{m} \]
Optimal radius:
\[ r^* \approx 11.28\ \text{m} \]

Habitat Values for Different Depths

Using the volume constraint
\[ r = \sqrt{\frac{200}{\pi h}}, \] we can compute the pond radius and the corresponding habitat value for several feasible depths.

Depth $h$ (m)	Radius $r = \sqrt{200/(\pi h)}$ (m)	Habitat $H = \pi r^2 - 4h$
0.5	11.28	398.0
1.0	7.98	196.0
2.0	5.64	92.0
3.0	4.61	54.67
4.0	3.99	34.0

Interpretation of the Table

As depth increases, radius must decrease to keep volume fixed.
Because habitat value heavily rewards surface area, habitat declines rapidly with increasing depth.
The maximum habitat value occurs at the minimum feasible depth $h = 0.5$, consistent with the analytic solution.

Interpretation

The model rewards surface area and penalizes depth, so making the pond as wide and shallow as allowed yields the highest habitat value.
Without the minimum‐depth constraint, the model would push toward $r \to \infty,\ h \to 0$.
The ecological requirement $h \ge 0.5$ creates a realistic maximum.

8. Wind Turbine Height for Maximum Power Output

A wind turbine’s power output depends on height $h$ (meters):

\[ P(h) = 20h - 0.2h^2. \]

Your Task

Find the optimal turbine height.
Confirm with second derivative.
Explain why turbines cannot just be infinitely tall.

Show Solution

\[ P'(h) = 20 - 0.4h. \]

Set to zero:

\[ 20 - 0.4h = 0 \Rightarrow h = 50 \text{ m}. \]

Second derivative:

\[ P''(h) = -0.4 < 0. \]

Interpretation:
Height increases wind speed, but increased tower mass, bending stresses, and inefficiencies reduce marginal gains.
Fifty meters is the balance point.

9. Coral Reef Restoration: Optimizing Transplant Density

Transplanted coral fragments grow and fuse into colonies, but overcrowding increases mortality.
Model:

\[ G(d) = 30d - d^2, \]

where $d$ is transplant density (fragments per m²).

Your Task

Find density maximizing growth.
Confirm with second derivative.
Interpret.

Show Solution

\[ G'(d) = 30 - 2d. \]

Set to zero:

\[ 30 - 2d = 0 \Rightarrow d = 15. \]

Second derivative:

\[ G''(d) = -2 < 0. \]

Interpretation:
Low densities slow reef formation; high densities cause shading and disease.
Fifteen fragments/m² is optimal.

10. Urban Green Roof: Minimizing Stormwater Runoff

Green roof substrate depth $z$ affects stormwater retention:

\[ R(z) = 15\sqrt{z} - z. \]

Your Task

Find depth maximizing retention.
Use the second derivative test.
Explain sustainability implications.

Show Solution

\[ R'(z) = \frac{15}{2\sqrt{z}} - 1. \]

Set to zero:

\[ \frac{15}{2\sqrt{z}} = 1 \Rightarrow \sqrt{z} = \frac{15}{2} \Rightarrow z = \frac{225}{4} = 56.25. \]

Second derivative:

\[ R''(z) = -\frac{15}{4}z^{-3/2} < 0. \]

Interpretation:
Very shallow roofs store little water; very deep ones drain slowly and add weight.
A depth of 56 cm balances retention and structural load.

11. Kelp Forest Restoration: Optimizing Mooring Spacing

Kelp restoration teams anchor kelp seedlings on ropes suspended from floating buoys.
If buoys are too close, shading reduces kelp growth; if too far, total canopy area decreases.

Let spacing between buoys be $d$ meters.
Kelp canopy productivity per anchor rope is modeled as:

\[ P(d) = 80d - 4d^2. \]

Your Task

Find the spacing $d$ that maximizes kelp productivity.
Confirm the result using the second derivative.
Explain the ecological tradeoff shaping this optimum.

Show Solution

Differentiate:

\[ P'(d) = 80 - 8d. \]

Set to zero:

\[ 80 - 8d = 0 \Rightarrow d = 10. \]

Second derivative:

\[ P''(d) = -8 < 0, \]

so 10 m gives a maximum.

Interpretation:
Very tight spacing creates shading and drag; very wide spacing limits canopy formation.
A 10-m spacing balances light availability with total canopy area.

12. Oyster Reef Design: Maximize Filtration Surface

A reef restoration team designs reef modules of radius $r$ (m).
Surface area available for oyster attachment is:

\[ A(r) = 100r - 2r^2. \]

This captures:

increasing habitat with module size, but
diminishing efficiency as very large modules reduce water flow inside the reef.

Your Task

Find radius $r$ maximizing oyster habitat area.
Confirm with second derivative.
Explain why very large modules reduce effectiveness.

Show Solution

Differentiate:

\[ A'(r) = 100 - 4r. \]

Set to zero:

\[ 100 - 4r = 0 \Rightarrow r = 25. \]

Second derivative:

\[ A''(r) = -4 < 0. \]

Thus the maximum occurs at 25 m.

Interpretation:
Large structures restrict internal circulation, reducing actual usable surface area.
The optimum balances habitat size with efficient water flow.

13. Coral Nursery Lines: Optimizing Fragment Spacing

Floating coral nurseries suspend corals along horizontal lines.
Growth per coral fragment is modeled by:

\[ G(s) = 60s - 3s^2, \]

where $s$ is spacing (cm).

Your Task

Find the spacing that maximizes coral growth.
Use the second derivative test.
Explain the biological reasoning behind the shape of the model.

Show Solution

Differentiate:

\[ G'(s) = 60 - 6s. \]

Set to zero:

\[ 60 - 6s = 0 \Rightarrow s = 10. \]

Second derivative:

\[ G''(s) = -6 < 0. \]

Thus coral growth is maximized at 10 cm spacing.

Interpretation:
Close spacing increases competition for light and water flow; wide spacing wastes nursery capacity.
Ten centimeters provides an efficient density for fragment growth.

14. Hatchery Release Strategy: Optimal Smolt Size

Fish survival after release increases with smolt size $m$, but hatchery cost and predation during grow-out both increase with size.

Survival benefit:

\[ S(m) = 200m - 5m^2. \]

First term: larger smolts survive ocean entry better
Second term: diminishing returns and increased accumulation of hatchery cost/risk

Your Task

Find the size $m$ (grams) maximizing survival benefit.
Confirm with second derivative.
Provide ecological interpretation.

Show Solution

Differentiate:

\[ S'(m) = 200 - 10m. \]

Set to zero:

\[ 200 - 10m = 0 \Rightarrow m = 20\ \text{g}. \]

Second derivative:

\[ S''(m) = -10 < 0. \]

Thus 20 g maximizes survival benefit.

Interpretation:
Fish released too small face marine predation; too large require costly rearing and encounter density-dependent issues in tanks.
The optimum balances hatchery cost with early marine survival probability.

Practice: Last year’s test questions

Problem 1: Optimization — Stormwater Design

A community center is installing a rectangular rain garden along the side of its building to capture and filter stormwater.
The garden will be bordered by stone edging along two lengths and one width.
The other width is against a wall and does not require edging.

The stone edging costs $4 per meter, and the center has a total budget of $400.

(a) Write an equation that relates the total cost of the edging to width $x$ and length $y$.
(b) Use your equation to express the area $A(x) = xy$ as a function of a single variable.
(c) What dimensions will maximize the area of the rain garden within the budget?

Click to show solution

(a) Cost equation

Edging is required along two lengths and one width, so the total edging length is
\[ 2y + x. \]

Edging costs $4 per meter and the budget is $400: \[ 4(2y + x) = 400. \]

Divide both sides by 4: \[ 2y + x = 100. \]

Solve for $y$: \[ y = \frac{100 - x}{2}. \]

(b) Area as a function of one variable

The area is $A(x) = xy$. Substitute the expression for $y$: \[ A(x) = x\left(\frac{100 - x}{2}\right) = 50x - \frac12 x^2. \]

This is a downward-opening quadratic.

(c) Dimensions that maximize the area

First derivative: \[ A'(x) = 50 - x. \]

Set equal to zero to find the critical point: \[ 50 - x = 0 \quad\Rightarrow\quad x = 50. \]

Second derivative: \[ A''(x) = -1, \] which is negative, so the graph is concave down everywhere.
Thus $x = 50$ gives a maximum.

Endpoint check

Valid widths must satisfy $0 < x < 100$.

\[ A(0) = 0, \qquad A(100) = 0. \]

Interior value: \[ A(50) = 50\cdot 50 - \frac12 (50)^2 = 1250. \]

The critical point has the largest area.

Dimensions

\[ y = \frac{100 - 50}{2} = 25. \]

Maximum-area garden:
- Width $x = 50$ m
- Length $y = 25$ m
- Area $= 1250\ \text{m}^2$

Problem 2: Optimization — Tree Spacing

In a reforestation project, a conservation team is trying to determine the optimal spacing between rows of newly planted trees.

If the spacing is too narrow, the forest becomes too dense for understory plants.
If the spacing is too wide, there’s not enough shade to support sensitive species.

The team models the biodiversity benefit by:

\[ B(w) = w^2 - 18w + 100 \]

where $w$ is the width (in meters) between tree rows.
To maintain accessibility and ecological integrity, spacing must be between 4 and 10 meters.

(a) What spacing should the team choose to maximize biodiversity benefit?

Click to show solution

Objective

The biodiversity benefit is modeled by: \[ B(w) = w^2 - 18w + 100. \]

Spacing must satisfy: \[ 4 \le w \le 10. \]

The goal is to find the value of $w$ in this interval that maximizes $B(w)$.

First derivative

\[ B'(w) = 2w - 18. \]

Set the derivative equal to zero:

\[ 2w - 18 = 0 \quad\Rightarrow\quad w = 9. \]

This is the only critical point.

Second derivative

\[ B''(w) = 2. \]

Since $B''(w)$ is positive, the function is concave up, meaning the critical point $w = 9$ is a minimum, not a maximum.

So the maximum must occur at one of the endpoints.

Endpoint check

Evaluate the function at the boundaries of the allowed interval:

\[ B(4) = 4^2 - 18(4) + 100 = 16 - 72 + 100 = 44, \]

\[ B(10) = 10^2 - 18(10) + 100 = 100 - 180 + 100 = 20. \]

Compare with the interior value:

\[ B(9) = 9^2 - 18(9) + 100 = 81 - 162 + 100 = 19. \]

Conclusion

The largest value is: \[ B(4) = 44. \]

The spacing that maximizes biodiversity benefit is
\[ \boxed{w = 4\ \text{meters}}. \]

Problem 3: Optimization — Urban Tree Buffer Design

A city is creating a tree-planted buffer strip along a major roadway to reduce noise, absorb rainfall, and improve air quality.

The environmental benefit score is modeled by:

\[ E(w) = -\frac13 w^3 \;-\; \frac12 w^2 \;+\; 6w \]

where:

$w$ = width of the buffer (meters)
$E(w)$ = environmental benefit score (unitless)

(a) What is the objective function in this problem?
(b) What are the constraints (if any)?
(c) Find the buffer width that maximizes environmental benefit.
(d) If there were more than one critical point, explain how you would choose between them.

Click to show solution

(a) Objective function

The goal is to maximize the environmental benefit score: \[ E(w) = -\frac13 w^3 - \frac12 w^2 + 6w. \]

(b) Constraints

No explicit real-world constraints are stated, so mathematically we assume: \[ w > 0. \]

If desired, the city could later impose practical limits (e.g., available land width), but none are required to solve the problem.

(c) Maximize the environmental benefit

First derivative: \[ E'(w) = -w^2 - w + 6. \]

Set equal to zero: \[ -w^2 - w + 6 = 0. \]

Multiply by $-1$: \[ w^2 + w - 6 = 0. \]

Factor: \[ (w + 3)(w - 2) = 0. \]

Critical points: \[ w = -3,\quad w = 2. \]

Since width must be positive, the only feasible candidate is: \[ w = 2. \]

Second derivative test

\[ E''(w) = -2w - 1. \]

Evaluate at $w = 2$: \[ E''(2) = -5, \] which is negative, so the function is concave down at this point.
Thus $w = 2$ gives a local maximum of the benefit score.

Endpoint check (conceptual)

Because the domain is $w > 0$ and the cubic term is negative,
\[ E(w) \to -\infty \quad \text{as } w \to \infty. \]

And at very small positive widths: \[ E(w) \approx 6w > 0, \] so the function increases from 0, reaches a maximum at $w = 2$, and then decreases forever.

Thus the only maximum is at $w = 2$.

(d) Choosing between multiple critical points

If several critical points were feasible, we would:

Check the second derivative at each one (concave down → local max).
Evaluate the function at all candidates and compare values.
Apply domain constraints (e.g., discard negative or non-physical widths).
If the domain were closed, also check endpoints.

This ensures we select the point with the largest environmental benefit within all constraints.

Final result

The environmental benefit is maximized at: \[ \boxed{w = 2\ \text{meters}}. \]

Problem 4: Optimization — Compost Bin Design

A community garden is building a square base open-top compost bin.
The bin must have volume of 2.0 m³.
The base costs $30 per square meter, and the sides cost $20 per square meter.

Let:

$l$ = side length of the square base (meters)
$h$ = height of the bin (meters)

(a) Write an equation for the cost in terms of the bin dimensions.
What is the objective function?

(b) What is the constraint equation?

(c) Use the constraint to eliminate a variable and express the objective function in one variable.

(d) Determine the optimal dimensions of the bin.
Justify your reasoning.

Click to show solution

(a) Objective function: cost

The bin has:

A square base of area $l^2$, costing $30 per m²
Four sides, each of area $l \cdot h$, costing $20 per m²

Total cost: \[ C = 30l^2 + 4(20lh) \]

Simplify: \[ C = 30l^2 + 80lh. \]

This is the objective function to minimize.

(b) Constraint equation (volume)

\[ l^2 h = 2. \]

Solve for $h$: \[ h = \frac{2}{l^2}. \]

(c) Eliminate a variable

Substitute into the cost equation:

\[ C(l) = 30l^2 + 80l\left(\frac{2}{l^2}\right) = 30l^2 + \frac{160}{l}. \]

This expresses cost in terms of a single variable.

(d) Find the optimal dimensions

First derivative: \[ C'(l) = 60l - \frac{160}{l^2}. \]

Set to zero: \[ 60l = \frac{160}{l^2} \]

\[ 60l^3 = 160 \]

\[ l^3 = \frac{8}{3} \]

\[ l = \left(\frac{8}{3}\right)^{1/3}. \]

Second derivative test

\[ C''(l) = 60 + \frac{320}{l^3}. \]

This is positive for all $l>0$, so the critical point is a minimum.

Height calculation

\[ h = \frac{2}{l^2}. \]

Using
\[ l = \left(\frac{8}{3}\right)^{1/3}, \] we get the symbolic height: \[ h = \frac{3^{2/3}}{2}. \]

Numerical approximation:

$3^{1/3} \approx 1.44225$
$3^{2/3} \approx 2.08008$
$h = \frac{3^{2/3}}{2} \approx \frac{2.08008}{2} \approx 1.04$

So: \[ h \approx 1.04\ \text{m}. \]

Final optimal dimensions

Side length:
\[ l = \left(\frac{8}{3}\right)^{1/3} \approx 1.39\ \text{m} \]
Height:
\[ h = \frac{3^{2/3}}{2} \approx 1.04\ \text{m} \]

These dimensions minimize construction cost while meeting the volume requirement.