- $w$ : the base width
- $l$ : the base length
- $h$ : the height
1. Constraints and Relationships
- Length-Width Relationship: The base length is $3$ times the base width.
$$
l=3 w
$$
- Volume Constraint: The volume $V$ must be $50 \mathrm{ft}^3$.
$$
\begin{gathered}
V=l w h \\\\
50=(3 w) w h \\\\
50=3 w^2 h \quad(\text { Equation } 1)
\end{gathered}
$$
2. Cost Function (Objective Function)
The total cost $C$ is the sum of the cost of the top/bottom material and the cost of the side material.
- Area of Top and Bottom: $2 \times(l \cdot w)=2(3 w-w)=6 w^2$
- Cost Rate: $810 / \mathrm{ft}^2$
- Cost for Top/Bottom: $10 \times\left(6 w^2\right)=60 w^2$
- Area of the Sides (Lateral Area): $2 \times(l \cdot h)+2 \times(w \cdot h)=2(3 w h)+2(w h)=8 u h$
- Cost Rate: $86 / \mathrm{ft}^2$
- Cost for Sides: $6 \times(8 w h)=48 w h$
The total cost function $C$ is:
$$
C=60 w^2+48 w h \quad \text { (Equation 2) }
$$
To find the minimum cost using differentiation, we must express the Cost function $C$ in terms of only one variable, $w$.
1. Solve Equation 1 for $h:$
$$
h=\frac{50}{3 w^2}
$$
2. Substitute this expression for $h$ into the Cost Equation 2:
$$
C(w)=60 w^2+48 w\left(\frac{50}{3 w^2}\right)
$$
3. Simplify the expression:
$$
\begin{aligned}
& C(w)=60 w^2+\frac{48 \times 50}{3 w} \\\\
& C(w)=60 w^2+\frac{2400}{3 w} \\\\
& C(w)=60 w^2+\frac{800}{w} \\\\
& C(w)=60 w^2+800 w^{-1}
\end{aligned}
$$
We find the minimum cost by setting the first derivative of the cost function equal to zero.
1. Find the derivative of $C(w)$ with respect to $w$ :
$$
\begin{gathered}
\frac{d C}{d w}=120 w-800 w^{-2} \\\\
\frac{d C}{d w}=120 w-\frac{800}{w^2}
\end{gathered}
$$
2. Set the derivative to zero and solve for $w$ (the critical point):
$$
\begin{gathered}
0=120 w-\frac{800}{w^2} \\\\
120 w=\frac{800}{w^2} \\\\
120 w^3=800 \\\\
w^3=\frac{800}{120} \\\\
w^3=\frac{80}{12} \\\\
w^3=\frac{20}{3} \\\\
w=\sqrt[3]{\frac{20}{3}} \approx 1.886 \text { feet }
\end{gathered}
$$
3. Confirm the minimum (Optional): The second derivative $\frac{d^2 C}{d w^2}=120+1600 w^{-3}$. Since $w$ must be positive, the second derivative is positive, confirming that this value of $w$ is a minimum.
\begin{gathered}
l=3 w \\\\
l=3 \sqrt[3]{\frac{20}{3}} \\\\
\mathbf{l}=\sqrt[3]{\frac{\mathbf{2 7} \times \mathbf{2 0}}{\mathbf{3}}}=\sqrt[3]{\mathbf{9} \times \mathbf{2 0}}=\sqrt[3]{\mathbf{1 8 0}} \mathrm{ft} \quad(\approx \mathbf{5 . 6 5 9 ~ f t}) \\\\
h=\frac{50}{3 w^2} \\\\
h=\frac{50}{3\left(\sqrt[3]{\frac{20}{3}}\right)^2} \\\\
h=\frac{50}{3\left(\frac{20}{3}\right)^{2 / 3}}(\approx \mathbf{4 . 7 1 5 ~ f t})
\end{gathered}
WIDTH $(w): \sqrt[3]{\frac{20}{3}} ~\text{feet} \approx \mathbf{1.89 ~feet}$
LENGTH $(l) : \sqrt[3]{180} ~\text{feet} \approx \mathbf{5.66 ~feet}$
HEIGHT $(h): \frac{50}{3\left(\sqrt[3]{\frac{21}{3}}\right)^2} ~\text{feet} \approx \mathbf{4.72 ~feet}$
Option A is correct
