A theorem states that one local extremum implies absolute extremum. Verify that the following function satisfies the conditions of this theorem on its domain. Then,
find the location and value of the absolute extremum guaranteed by the theorem.

The function has an absolute
(Type exact answers, using as needed.)

Ask by Haynes Stewart. in the United States

Mar 28,2025

Question

A theorem states that one local extremum implies absolute extremum. Verify that the following function satisfies the conditions of this theorem on its domain. Then,
find the location and value of the absolute extremum guaranteed by the theorem.

The function has an absolute
(Type exact answers, using as needed.)

Ask by Haynes Stewart. in the United States

Mar 28,2025

UpStudy AI · Accepted Answer

We are given
$$
A(r)=\frac{40}{r}+4\pi r^2,\quad r>0.
$$

**Step 1. Find the derivative**

Differentiate with respect to $$ r $$:
$$
A'(r)= -\frac{40}{r^2}+8\pi r.
$$

**Step 2. Find the critical point**

Set the derivative equal to zero:
$$
-\frac{40}{r^2}+8\pi r =0.
$$
Solve for $$ r $$:
$$
8\pi r =\frac{40}{r^2} \quad\Longrightarrow\quad 8\pi r^3=40.
$$
Thus,
$$
r^3=\frac{40}{8\pi}=\frac{5}{\pi}\quad\Longrightarrow\quad r=\sqrt[3]{\frac{5}{\pi}}.
$$

**Step 3. Check that the critical point is a minimum**

Compute the second derivative:
$$
A''(r)=\frac{d}{dr}\left(-\frac{40}{r^2}+8\pi r\right)=\frac{80}{r^3}+8\pi.
$$
Since $$ r>0 $$, both terms are positive; hence, $$ A''(r)>0 $$. This shows that the function is convex and that the critical point is a local minimum. Moreover, because
- as $$ r\to 0^+,\ \frac{40}{r}\to \infty $$, and 
- as $$ r\to \infty,\ 4\pi r^2\to \infty $$,
this local minimum is also the absolute minimum.

**Step 4. Find the value of the function at the minimum**

Substitute $$ r=\sqrt[3]{\frac{5}{\pi}} $$ into $$ A(r) $$:
$$
A\Bigl(\sqrt[3]{\frac{5}{\pi}}\Bigr)=\frac{40}{\sqrt[3]{\frac{5}{\pi}}}+4\pi\left(\sqrt[3]{\frac{5}{\pi}}\right)^2.
$$
Rewrite the terms:
$$
\frac{40}{\sqrt[3]{\frac{5}{\pi}}}=40\left(\frac{\pi}{5}\right)^{\frac{1}{3}}
$$
and
$$
4\pi\left(\sqrt[3]{\frac{5}{\pi}}\right)^2=4\pi\left(\frac{5}{\pi}\right)^{\frac{2}{3}}=4\cdot5^{\frac{2}{3}}\pi^{1-\frac{2}{3}}=4\cdot5^{\frac{2}{3}}\pi^{\frac{1}{3}}.
$$
Thus,
$$
A\Bigl(\sqrt[3]{\frac{5}{\pi}}\Bigr)=40\pi^{\frac{1}{3}}5^{-\frac{1}{3}}+4\pi^{\frac{1}{3}}5^{\frac{2}{3}}
=\pi^{\frac{1}{3}}\left(40\cdot5^{-\frac{1}{3}}+4\cdot5^{\frac{2}{3}}\right).
$$

Notice that
$$
40\cdot5^{-\frac{1}{3}}=8\cdot5^{\frac{2}{3}},
$$
since multiplying $$ 8\cdot5^{\frac{2}{3}} $$ by $$ 5^{\frac{1}{3}} $$ gives $$ 8\cdot5=40 $$. Therefore,
$$
40\cdot5^{-\frac{1}{3}}+4\cdot5^{\frac{2}{3}}=8\cdot5^{\frac{2}{3}}+4\cdot5^{\frac{2}{3}}=12\cdot5^{\frac{2}{3}}.
$$
So the absolute minimum value is
$$
A\Bigl(\sqrt[3]{\frac{5}{\pi}}\Bigr)=12\cdot5^{\frac{2}{3}}\pi^{\frac{1}{3}}.
$$

**Final Answer**

The function has an absolute minimum at
$$
r=\sqrt[3]{\frac{5}{\pi}},
$$
with the minimum value
$$
A(r)=12\cdot5^{\frac{2}{3}}\pi^{\frac{1}{3}}.
$$
The function $$ A(r) = \frac{40}{r} + 4\pi r^2 $$ has an absolute minimum at $$ r = \sqrt[3]{\frac{5}{\pi}} $$, with the minimum value $$ A(r) = 12 \cdot 5^{\frac{2}{3}} \pi^{\frac{1}{3}} $$.

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