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

Explanation:

Step1: Find the derivative of A(r)

Differentiate $A(r)=\frac{36}{r}+3\pi r^{2}$ with respect to $r$. Using the power - rule $\frac{d}{dr}(x^n)=nx^{n - 1}$, we have $A'(r)=-\frac{36}{r^{2}}+6\pi r$.

Step2: Set the derivative equal to zero

Set $A'(r) = 0$, so $-\frac{36}{r^{2}}+6\pi r=0$. Multiply through by $r^{2}$ to get $- 36+6\pi r^{3}=0$. Then $6\pi r^{3}=36$, and $r^{3}=\frac{6}{\pi}$, so $r = \sqrt[3]{\frac{6}{\pi}}$.

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

Differentiate $A'(r)=-\frac{36}{r^{2}}+6\pi r$ with respect to $r$. $A''(r)=\frac{72}{r^{3}}+6\pi$.

Step4: Evaluate the second - derivative at the critical point

Substitute $r = \sqrt[3]{\frac{6}{\pi}}$ into $A''(r)$. $A''(\sqrt[3]{\frac{6}{\pi}})=\frac{72}{\frac{6}{\pi}}+6\pi=12\pi + 6\pi=18\pi>0$. So the function has a local minimum at $r=\sqrt[3]{\frac{6}{\pi}}$.

Step5: Find the absolute minimum value

Substitute $r=\sqrt[3]{\frac{6}{\pi}}$ into $A(r)$. [ \begin{align*} A(\sqrt[3]{\frac{6}{\pi}})&=\frac{36}{\sqrt[3]{\frac{6}{\pi}}}+3\pi(\sqrt[3]{\frac{6}{\pi}})^{2}\ &=36(\frac{\pi}{6})^{\frac{1}{3}}+3\pi(\frac{6}{\pi})^{\frac{2}{3}}\ &=36(\frac{\pi}{6})^{\frac{1}{3}} + 3\pi\times\frac{6^{\frac{2}{3}}}{\pi^{\frac{2}{3}}}\ &=36(\frac{\pi}{6})^{\frac{1}{3}}+ 18(\frac{6}{\pi})^{\frac{1}{3}}\ &=9(\frac{\pi}{6})^{\frac{1}{3}}+9(\frac{\pi}{6})^{\frac{1}{3}}+ 18(\frac{6}{\pi})^{\frac{1}{3}}\ &=9(\frac{\pi}{6})^{\frac{1}{3}}+9(\frac{\pi}{6})^{\frac{1}{3}}+9(\frac{\pi}{6})^{\frac{1}{3}}+9(\frac{\pi}{6})^{\frac{1}{3}}\ &= 9\left[(\frac{\pi}{6})^{\frac{1}{3}}+(\frac{\pi}{6})^{\frac{1}{3}}+(\frac{6}{\pi})^{\frac{1}{3}}+(\frac{6}{\pi})^{\frac{1}{3}}\right]\ &=9\left[2(\frac{\pi}{6})^{\frac{1}{3}}+2(\frac{6}{\pi})^{\frac{1}{3}}\right]\ &=18\left[(\frac{\pi}{6})^{\frac{1}{3}}+(\frac{6}{\pi})^{\frac{1}{3}}\right] \end{align*} ]

Answer:

The function $A(r)=\frac{36}{r}+3\pi r^{2},r > 0$ has an absolute extremum of $18\left(\left(\frac{\pi}{6}\right)^{\frac{1}{3}}+\left(\frac{6}{\pi}\right)^{\frac{1}{3}}\right)$ at $r=\sqrt[3]{\frac{6}{\pi}}$.