Application: Optimization II

Problem 230: A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $40/ft and on the other three sides by a metal fence costing $20/ft. If the area of the garden is 182 square feet, find the dimensions of the garden that minimize the cost.

Solution: Let $x$ be the length of the brick wall and $y$ be the length of an adjacent side with $x,y \ge 0$ . The area is given by

(1) $\begin{equation*} A = x \cdot y \implies 182 = x \cdot y \implies y = \frac{182}{x}.\end{equation*}$

The cost if given by

(2) $\begin{align*} \text{Cost} = C(x) & = 40x + 20(2y + x) \\ & = 40x + 20 \left( \frac{2(182)}{x}+x\right) \\ & = 40x + 20 \left( \frac{364}{x} + x \right) \\ & = 40x + \frac{7,280}{x} + 20 x \\ & = 60x + \frac{7,280}{x}. \end{align*}$

We now take the derivative and set it equal to zero to find critical points.

(3) $\begin{align*}C(x) & = 60x + \frac{7,280}{x} \\ C'(x) & = 60 - \frac{7,280}{x^2} \\ 0 & = 60 - \frac{7,280}{x^2} \\ 60x^2 & = 7,280 \\ x & = \sqrt{7,280/60} \\ & = 11.0151410946. \end{align*}$

Using $x$ we can find $y$ .

(4) $\begin{equation*} y = \frac{182}{x} = \frac{182}{11.011410946} = 16.5227116419. \end{equation*}$

Thus to minimize the cost we need $x \approx 11.02$ length of the brick wall and $y \approx 16.52$ length of the adjacent side.

Leave a Reply Cancel reply