Question

A rancher has 400 feet of fencing to put around a rectangular field and then subdivide the field into 2 identical smaller rectangular plots by placing a fence parallel to one of the field's shorter sides. find the dimensions that maximize the enclosed area. write your answers as fractions reduced to lowest terms.

Answer 1

Let's assume

length of rectangle is x

width of rectangle is y

now, we can draw rectangle

now, we can find fencing length

$F=2x+3y$

we are given

total fencing length=400

so, we can set them equal

$400=2x+3y$

now, we can solve for y

$3y=400-2x$

$y=\frac{400}{3} -\frac{2x}{3}$

now, we can find area

Area=length*width

$A=xy$

we can plug back y

$A=x(\frac{400}{3} -\frac{2x}{3})$

$A=\frac{400}{3}x -\frac{2x^2}{3}$

we have to maximise area

For maximising area , firstly we find derivative of area

and then we can set derivative =0 and then we solve for x

so, firstly we will find derivative

$A'=\frac{400}{3}*1 -\frac{2*2x}{3}$

now, we can set it to 0

and then we solve for x

$0=\frac{400}{3}*1 -\frac{2*2x}{3}$

$400-4x=0$

$x=100$

now, we can find y

$y=\frac{400}{3} -\frac{2x}{3}$

$y=\frac{400}{3} -\frac{2*100}{3}$

$y=\frac{200}{3}$

so, dimensions are

length is 100 feet

width is $\frac{200}{3}$ feet.............Answer