Question

A farmer has 200 feet to fence to

enclose a regular area for goats. What

dimensions for the rectangle will maximize the area for the goats?

Answer 1

Answer:

The dimensions for the rectangle are  'x'= 50ft and 'y'= 50ft

The maximum area for the goats = 2500ft^2

Step-by-step explanation:

Step(l):-

Given a farmer has 200 feet to fence to   enclose a regular area for goats

Given perimeter of the rectangle

                                         2(x + y) = 200

                                           x + y = 100

                                            y = 100 -x …(l)

Step(ll):-

The Area of the rectangle = length × breadth

                                  A = x y

                                 A = x (100 -x)

                                A = 100x - x² …(ii)

Differentiating equation (ii) with respective to 'x' we get

                              $\frac{dA}{dx} = 100 -2x …(lll)$

                             $\frac{dA}{dx} = 100 -2x =0$

                            100 -2x =0

                            100 = 2x

                                 $x = \frac{100}{2} = 50$

Step(lll):-

Again  differentiating equation (lll) with respective to 'x' , we get

$\frac{d^{2} A}{dx^{2} } = -2 < 0$

The maximum value at x = 50

                                     y = 100 - x

                                     y = 100 - 50 =50

The dimensions are x = 50 and y = 50

The Area of the rectangle = 50×50 = 2500ft^2

Conclusion:-

The dimensions for the rectangle

length of the rectangle 'x'= 50ft and  

width of the rectangle 'y'= 50ft

The maximum area for the goats = 2500ft^2