Question

A farmer is building a fence to enclose a rectangular area consisting of two separate regions. the four walls and one additional vertical segment (to separate the regions) are made up of fencing, as shown below. a dashed and shaded rectangle is cut into two equal halves by a dashed vertical line. a dashed and shaded rectangle is cut into two equal halves by a dashed vertical line. if the farmer has 648 feet of fencing, what are the dimensions of the region which enclose the maximal area

Answer 1

The dimensions of the region which enclose the maximal area will be:  Length = 162 feet and Width = 108 feet.

Explanation

According to the below diagram, the total rectangular area is divided into two separate regions by a vertical segment.

Suppose, the length and width of the rectangular area are  L and W respectively.

So, the length of that vertical segment will be equal to the width, W.

If the four walls of the rectangular area and the vertical segment are made up of fencing, then the total fence required $=(2L+3W)$ feet.

Given that, the farmer has 648 feet of fencing. So, the equation will be .....

$2L+3W= 648\\ \\ 2L= 648-3W\\ \\ L=324-\frac{3}{2}W .......................... (1)$

Now, the area of the rectangular area:  $A= L*W ...................................... (2)$

Substituting equation (1) into equation (2) , we will get ......

$A=(324-\frac{3}{2}W)*W\\ \\ A= 324W- \frac{3}{2}W^2$

Taking derivative on the both sides of the above equation with respect to W, we will get ......

$\frac{dA}{dW}= 324-\frac{3}{2}(2W)\\ \\ \frac{dA}{dW}= 324-3W$

Now, A will be maximum when $\frac{dA}{dW}= 0$. So....

$324-3W=0\\ \\ 3W=324\\ \\ W=\frac{324}{3}=108$

Plugging this $W=108$ into equation (1) ........

$L=324-\frac{3}{2}(108)=324-162=162$

So, the dimensions of the region which enclose the maximal area will be:  Length = 162 feet and Width = 108 feet.

Answer 2

The dimensions that maximize the area are 108 ft by 162 ft

How to determine the dimension of the region?

Represent the dimension with x (width) and y (length)

The rectangular area is divided into two sections.

So, the area (A) and the perimeter (P) are:

$A = xy$

$P= 2y + 3x$

The perimeter is 648.

So, we have:

$2y + 3x = 648$

Make x the subject

$x = \frac{648 - 2y}3$

Substitute $x = \frac{648 - 2y}3$ in $A = xy$

$A = \frac{648 - 2y}{3} * y$

Evaluate the product

$A = \frac{648y - 2y^2}{3}$

Expand

$A = \frac{648y}{3} - \frac{2y^2}{3}$

Differentiate

$A' = \frac{648}{3} - \frac{4y}{3}$

Set to 0

$\frac{648}{3} - \frac{4y}{3} = 0$

Multiply through by 3

$648- 4y =0$

Divide through by 4

$162- y =0$

Solve for y

$y = 162$

Recall that: $x = \frac{648 - 2y}3$

So, we have:

$x = \frac{648 - 2*162}3$

Evaluate

$x = 108$

Hence, the dimensions that maximize the area are 108 ft by 162 ft

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