Question

Farmer Ed has 300meters of​ fencing, and wants to enclose a rectangular plot that borders on a river. If Farmer Ed does not fence the side along the​ river, find the length and width of the plot that will maximize the area. What is the largest area that can be​ enclosed?
Width
Length
Largest area:

Answer 1

L+2w=300

A=(w)(-2w+320)
A=-2w^2 +320w

A=-2(w^2 - 160w)
A=-2(w^2 - 160w+6400) +12800

A=-2(w-80)^2 +12800

The max area is 12800 m squared.

The width is 80 m and the length is 160m.

Answer 2

Answer:

Step-by-step explanation:

Alright, lets get started.

Suppose the width is W.

Suppose the length is L.

Total fence is given as 300, so

$L+W+W=300$

$L+2W=300$......................(1)

$Area=Length*Width$

$Area=L*W$

Plugging the value of L from equation (1)

$Area=(300-2W)*W$

$Area=300W-2W^2$

For Area to be maximized, derivative of area should be equal to zero.

Means :

$\frac{d}{dW}Area= 300-4W = 0$

So, $So, W Width = \frac{300}{4}=75$

So, $L Length=300-2*75=150$

So, Largest area= $75*150=11250$   :    Answer