# A rectangular plot of land is to be fenced in using two kinds of fencing. Two opposite sides will use fencing selling \$5 a foot.....while the remaining two sides use fencing selling \$2 a...

A rectangular plot of land is to be fenced in using two kinds of fencing. Two opposite sides will use fencing selling \$5 a foot..

...while the remaining two sides use fencing selling \$2 a foot. What are the dimensions of the greatest area that can be fenced in at a cost of \$4000?

lemjay | High School Teacher | (Level 2) Senior Educator

Posted on

Let x and y be the sides of the rectangle.

Then, set-up the equation for the total cost of the fence. To do so, let's use a \$5 per foot fencing for the two sides "x" . And, \$2 per foot fencing for the two sides "y".

So the equation is:

`4000 = 5(2x) + 2(2y)`

`4000=10x+4y`

To simplify, divide both sides by its GCF which is 2.

`4000/2=(10x+4y)/2`

`2000 =5x + 2y`

Then, solve for y.

`2000-5x=2y`

`(2000-5x)/2= y`

`1000-5/2x=y`

Since we have to maximize the area of the rectangle, set-up the equation of area.

`A =xy`

Note that `y= 1000-5/2x` . So,

`A=x(1000-5/2x)`

`A=1000x-5/2x^2`

Then, take the derivative of A.

`A' =1000-5x`

Set A' equal to zero and solve for x.

`0 = 1000-5x`

`5x=1000`

`x = 200`

Substitute the value of x to `y = 1000-5/2x` .

`y=1000-5/2*200 = 1000-500=500`

Hence, the dimensions of the rectangular plot that would maximize its area, given the total cost of fencing, is `200 xx 500` ft.