a theorem states that one local extremum implies absolute extremum. verify that the following function…

Explanation:

Step1: Find the derivative of A(r)

We have $A(r)=\frac{50}{r}+5\pi r^{2}=50r^{- 1}+5\pi r^{2}$. Using the power - rule $(x^n)^\prime=nx^{n - 1}$, the derivative $A^\prime(r)=-50r^{-2}+10\pi r=\frac{-50}{r^{2}} + 10\pi r$.

Step2: Set the derivative equal to zero

Set $A^\prime(r) = 0$, so $\frac{-50}{r^{2}}+10\pi r = 0$. Multiply through by $r^{2}$ (since $r>0$) to get $-50 + 10\pi r^{3}=0$. Then $10\pi r^{3}=50$, and $r^{3}=\frac{5}{\pi}$, so $r=\sqrt[3]{\frac{5}{\pi}}$.

Step3: Find the second - derivative of A(r)

Differentiate $A^\prime(r)=\frac{-50}{r^{2}}+10\pi r=-50r^{-2}+10\pi r$ with respect to $r$. Using the power - rule, $A^{\prime\prime}(r)=100r^{-3}+10\pi=\frac{100}{r^{3}}+10\pi$. Substitute $r = \sqrt[3]{\frac{5}{\pi}}$ into $A^{\prime\prime}(r)$: $A^{\prime\prime}(\sqrt[3]{\frac{5}{\pi}})=\frac{100}{\frac{5}{\pi}}+10\pi=20\pi + 10\pi=30\pi>0$. So $r=\sqrt[3]{\frac{5}{\pi}}$ is a local minimum.

Step4: Find the value of the absolute minimum

Substitute $r=\sqrt[3]{\frac{5}{\pi}}$ into $A(r)$: [ \begin{align*} A(\sqrt[3]{\frac{5}{\pi}})&=\frac{50}{\sqrt[3]{\frac{5}{\pi}}}+5\pi(\sqrt[3]{\frac{5}{\pi}})^{2}\ &=50(\frac{\pi}{5})^{\frac{1}{3}}+5\pi(\frac{5}{\pi})^{\frac{2}{3}}\ &=50(\frac{\pi}{5})^{\frac{1}{3}}+5\pi\times\frac{5^{\frac{2}{3}}}{\pi^{\frac{2}{3}}}\ &=50(\frac{\pi}{5})^{\frac{1}{3}}+25(\frac{\pi}{5})^{\frac{1}{3}}\ &=75(\frac{\pi}{5})^{\frac{1}{3}} \end{align*} ]

Answer:

The function $A(r)=\frac{50}{r}+5\pi r^{2},r > 0$ has an absolute extremum of $75(\frac{\pi}{5})^{\frac{1}{3}}$ at $r=\sqrt[3]{\frac{5}{\pi}}$.