Answer:
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'
       [tex]\frac{dA}{dx} = 100 (1) - 2 (2x)[/tex]
   ⇒   [tex]\frac{dA}{dx} = 100 - 4 x[/tex]    ...(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
  [tex]\frac{d^{2} A}{dx^{2} } = -4 (1) < 0[/tex]
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
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