what is the maximum vertical distance between the line ( y = x + 72 ) and the parabola ( y = x^{2} ) for (…

Explanation:

Step1: Define the distance function

Let (d(x)=(x + 72)-x^{2}), since the vertical distance between (y=x + 72) and (y = x^{2}) is (d(x)=upper - lower) function. So (d(x)=-x^{2}+x + 72).

Step2: Find the derivative of the distance function

Differentiate (d(x)) with respect to (x). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (d^\prime(x)=-2x+1).

Step3: Find the critical points

Set (d^\prime(x) = 0), then (-2x + 1=0). Solving for (x) gives (x=\frac{1}{2}).

Step4: Evaluate the distance function at the critical point and endpoints

• Evaluate (d(x)) at (x=-8): (d(-8)=-(-8)^{2}+(-8)+72=-64 - 8+72=0).
• Evaluate (d(x)) at (x=\frac{1}{2}): (d(\frac{1}{2})=-(\frac{1}{2})^{2}+\frac{1}{2}+72=-\frac{1}{4}+\frac{1}{2}+72=\frac{-1 + 2}{4}+72=\frac{1}{4}+72=72.25).
• Evaluate (d(x)) at (x = 9): (d(9)=-(9)^{2}+9+72=-81+9+72=0).

Answer:

(72.25)