Question

there are $400 available to fence in a rectangular garden. the fencing for the side of the garden facing the road costs $15 per foot, and the fencing for the other three sides costs $5 per foot. the picture on the right depicts this situation. consider the problem of finding the dimensions of the largest possible garden.
(a) determine the objective and constraint equations.
(b) express the quantity to be maximized as a function of x.
(c) find the optimal values of x and y
(a) let a equal the area of the garden. the objective equation is
(type an equation )

Explanation:

Step1: Define variables and objective

The area of a rectangle is length times width. Let the side along the road be $x$ and the other - side be $y$. The objective is to maximize the area $A$ of the rectangular garden. So the objective equation is $A=xy$.

Step2: Determine the constraint equation

The cost of the fencing is given. The cost of the side facing the road (length $x$) is $15x$ and the cost of the other three sides (two sides of length $y$ and one side of length $x$) is $5x + 5y+5y=5x + 10y$. The total budget for fencing is $400$. So the constraint equation is $15x+5x + 10y=400$, which simplifies to $20x + 10y=400$ or $y = 40 - 2x$.

Step3: Express the area as a function of $x$

Substitute $y = 40 - 2x$ into the area formula $A=xy$. We get $A(x)=x(40 - 2x)=40x-2x^{2}$.

Step4: Find the maximum of the function

To find the maximum of the function $A(x)=40x - 2x^{2}$, we take the first - derivative. $A^\prime(x)=\frac{d}{dx}(40x - 2x^{2})=40 - 4x$.
Set $A^\prime(x)=0$ to find the critical points. So $40 - 4x = 0$, which gives $x = 10$.
Then we take the second - derivative $A^{\prime\prime}(x)=\frac{d}{dx}(40 - 4x)=-4<0$. Since $A^{\prime\prime}(x)<0$ when $x = 10$, the function $A(x)$ has a maximum at $x = 10$.
Substitute $x = 10$ into the equation $y = 40 - 2x$ to find $y$. $y=40-2\times10 = 20$.

Answer:

(a) Objective equation: $A = xy$, Constraint equation: $20x + 10y=400$
(b) $A(x)=40x - 2x^{2}$
(c) $x = 10$, $y = 20$