Questions: 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.

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.

Transcript text: 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 $y$.

Solution

Solution Steps

Step 1: Define the Objective and Constraint Equations

Let $ x $ be the length of the side facing the road.
Let $ y $ be the length of the other side.
The cost for the side facing the road is $18 per foot, and the cost for the other three sides is $6 per foot.

The total cost equation (constraint) is: \[ 18x + 6(2y + x) = 384 \]

Simplify the constraint equation: \[ 18x + 6(2y + x) = 384 \] \[ 18x + 12y + 6x = 384 \] \[ 24x + 12y = 384 \] \[ 2x + y = 32 \]

The objective is to maximize the area $ A $: \[ A = x \cdot y \]

Step 2: Express the Quantity to be Maximized as a Function of $ x $

From the constraint equation $ 2x + y = 32 $, solve for $ y $: \[ y = 32 - 2x \]

Substitute $ y $ into the area equation: \[ A = x \cdot (32 - 2x) \] \[ A = 32x - 2x^2 \]

Step 3: Find the Optimal Values of $ x $ and $ y $

To find the maximum area, take the derivative of $ A $ with respect to $ x $ and set it to zero: \[ \frac{dA}{dx} = 32 - 4x \] Set the derivative equal to zero: \[ 32 - 4x = 0 \] \[ 4x = 32 \] \[ x = 8 \]

Substitute $ x = 8 $ back into the constraint equation to find $ y $: \[ y = 32 - 2(8) \] \[ y = 16 \]