A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic function A(x)=x(100−2x) gives the area, A, of the dog run for the length, x, of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.

a) The length of the building that should border the dog run to give the maximum area = 25feet

b)    The maximum area of the dog run  = 1250 s q feet²

Step-by-step explanation:

Step(i):-

Given function

                       A(x) = x (100-2x)

                      A (x) = 100x - 2x²...(i)

Differentiating equation (i) with respective to 'x'

     ⇒          ...(ii)

Equating  zero

         ⇒ 100 - 4x =0

         ⇒  100 = 4x

Dividing '4' on both sides , we get

             x = 25

Step(ii):-

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

Therefore The maximum value at x = 25

The length of the building that should border the dog run to give the maximum area = 25

Step(iii)

  Given  A (x) = x ( 100 -2 x)

substitute  'x' = 25 feet

             A(x) = 25 ( 100 - 2(25))

                    = 25(50)

                   = 1250

Conclusion:-

   The maximum area of the dog run  = 12 50  s q feet²

the ratio of balloons to tables is 5: 2 and the constant variation is 5.

the equation is y=5x

explanation:

step-by-step explanation: