Question

A farmer is planning a rectangular area for her chickens. The area of the rectangle will be 800 square feet. Three sides of the rectangle will be formed by fencing, which costs $5 per foot. The fourth side of the rectangle will be formed by a portion of the barn wall, which requires no fencing. In order to minimize the cost of the fencing, how long should the fourth side be?

Answer 1

Answer:

  40 ft

Step-by-step explanation:

Let x represent the length in feet of the fourth side. Then the sides perpendicular to the barn wall will have length 800/x, and the total cost of the fence will be ...

  cost = $5 × (x + 2·800/x)

The derivative of cost with respect to x will be zero when the cost is a minimum:

  d(cost)/dx = 5 -8000/x^2 = 0

  5x^2 = 8000 . . . . . multiply by x^2, add 8000

  x = √1600 =  40 . . . . feet . . . . . . divide by 5, take the square root

The length of the fourth side should be 40 feet.

Answer 2

Answer:

40 feet.

Step-by-step explanation:

Area of rectangle =  800 square feet

Let the fourth side of the rectangle be x.

Length of the rectangle = x feet

The area of a rectangle is

$A=length\times width$

$800=x\times width$

$\frac{800}{x}=width$

Cost of fencing = $5 per foot

Cost function on three sides is

$C(x)=5[length+2(width)]$

$C(x)=5x+10(\frac{800}{x})$

$C(x)=5x+\frac{8000}{x}$

Differentiate with respect to x.

$C'(x)=5-\frac{8000}{x^2}$           $[\because \frac{d}{dx}(\frac{1}{x})=-\frac{1}{x^2}]$

To find the critical point equate C'(x)=0.

$0=5-\frac{8000}{x^2}$

$-5=-\frac{8000}{x^2}$

$-5x^2=-8000$

Divide both sides by -5.

$x^2=-\frac{8000}{-5}$

$x^2=1600$

Taking square root on both sides,

$x=40$

Differentiate C'(x) with respect to x.

$C''(x)=\frac{16000}{x^3}>0$

Since C''(x)>0 for x=40, therefore cost of the fencing is minimum at x=40.

Thus, the measure of fourth side of the rectangle is 40 feet.