there are $384 available to fence in a rectangular garden. the fencing for the side of the garden facing the…

there are $384 available to fence in a rectangular garden. the fencing for the side of the garden facing the road costs $18 per foot, and the fencing for the other three sides costs $6 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.) the constraint equation is . (type an equation.) (b) express the quantity to be maximized as a function of x. (type an equation.) (c) find the optimal values of x and y. x = y =

there are $384 available to fence in a rectangular garden. the fencing for the side of the garden facing the road costs $18 per foot, and the fencing for the other three sides costs $6 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.) the constraint equation is . (type an equation.) (b) express the quantity to be maximized as a function of x. (type an equation.) (c) find the optimal values of x and y. x = y =

Answer

Explanation:

Step1: Find objective equation

The area of a rectangle is length times width. So the objective equation (area $A$) is $A = xy$.

Step2: Find constraint equation

The cost for the side facing the road is $18x$ and for the other three sides is $6x + 6y+6y=6x + 12y$. The total budget is $384$. So the constraint equation is $18x+6x + 12y=384$, which simplifies to $24x+12y = 384$ or $y=\frac{384 - 24x}{12}=32 - 2x$.

Step3: Express quantity to be maximized as function of x

Substitute $y = 32 - 2x$ into the area formula. So $A(x)=x(32 - 2x)=32x-2x^{2}$.

Step4: Find optimal values

Take the derivative of $A(x)$: $A'(x)=32 - 4x$. Set $A'(x)=0$ for critical - points. So $32 - 4x = 0$, which gives $x = 8$. Substitute $x = 8$ into $y=32 - 2x$, we get $y=32-2\times8 = 16$.

Answer:

(a) Objective equation: $A = xy$; Constraint equation: $24x + 12y=384$ (b) $A(x)=32x - 2x^{2}$ (c) $x = 8$, $y = 16$