Yucation
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.

Length, x =25 ft

Max area, A = 1250 ft square

Step-by-step explanation:

A(x) = x(100 – 2x)

       = 100x - 2x^2

To find the maximum value of A, we first find derivative of function of A to x

dA/dx = 100 - 4x

Find the critical value by dA/dx = 0

dA/dx = 100 - 4x = 0

100 - 4x = 0

4x = 100

Hence, the critical point x = 100/4 = 25

Find 2nd derivative to check if the equation has max value

d(dA)/dx^2 = -4

2nd derivative is negative, hence have maximum value

maximum value in this case is when the value of x = 25

The maximum area is therefore (substituting x = 25 into equation of A)

A = 25(100 - 2(25))

   = 25(50)

   = 1250 ft^2

10cm

step-by-step explanation:

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