Question

Farmer Ed has 800 meters 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?

Answer 1

Answer:

Therefore the dimension of the plot is 400 m × 200 m.

The area of the plot is = 80000 m²

Step-by-step explanation:

Given that a farmer ED has 800 meters of fencing, and wants to enclose a rectangular plot that borders on a river.

Let the length of side along river side be L.

And the width of the plot be be W.

Since the farmer does not fence the river side.

So the length of fence = (L+2w) m

According to the problem,

(L+2w) = 800

⇒L=800 - 2w

Therefore the area of the plot is A = Length ×width

                                                         = (L×W) m²

                                                        =[(800-2W)W] m²

For maximum or minimum $\frac{dA}{dW}=0$

∴ A =(800-2W)W

⇒A = 800 W -2W²

Differentiating with respect to W

$\frac{dA}{dW} = 800 -2.(2W)$

$\Rightarrow \frac{dA}{dW} = 800 -4W$

For maximum or minimum $\frac{dA}{dW}=0$

$\Rightarrow 0 = 800 -4W$

$\Rightarrow W= \frac{800}{4}$

⇒W=200

Again,

$\frac{dA}{dW} = 800 -4W$

Again differentiating with respect to W

$\frac{d^2A}{dW^2}= -4$

$\therefore[ \frac{d^2A}{dW^2}]_{W=200}= -4$

When $\frac{d^2A}{dW^2}$ , then the area A is maximum at W=200.

When $\frac{d^2A}{dW^2}>0$ , then the area A is minimum at W=200.

Therefore at W= 200 m , the area of the plot maximum at   W=200m

The length of the plot is L = [800- (2×200)] = 400 m

Therefore the dimension of the plot is 400 m × 200 m.

The area of the plot is = (400× 200) m²

                                      = 80000 m²