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What's the maximum volume a right circular cone with a slant height of 45 length units possible can have?

The volume $V$ for a right circular cone is given by $V = \frac{\pi r^2 h}{3}$.

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  • $\begingroup$ @ClaudeLeibovici Cone. Corrected it. $\endgroup$ – Markus Mar 25 '15 at 8:40
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Hint

Starting from lab bhattacharjee's answer, $$V(h)=\dfrac{\pi}3h(45^2-h^2)$$ and you want to maximize $V$.

When is reached an extremum of a function ? So, since you know it, write the corresponding condition and solve for $h$. Plug this value in $V(h)$ and use the second derivative test to confirm that the result corresponds to a maximum.

I am sure that you can take from here.

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We have $r^2+h^2=45^2\iff r^2=\cdots$

$$V=\dfrac{\pi r^2h}3=\dfrac{\pi}3h(45^2-h^2)=f(h)$$

Apply Second Derivative Test on $f(h)$

or $h(45^2-h^2)=\dfrac{h(45+h)(90-2h)}2\le\dfrac12\left(\dfrac{h+45+h+2(45-h)}3\right)^3=\dfrac{45^3}2$

as AM-GM inequality says: $\dfrac{a+b+c}3\ge\sqrt[3]{abc}\implies\left(\dfrac{a+b+c}3\right)^3\ge abc$

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