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I want to find the maximum volume block with dimensions $a$, $b$ and $c$ incribed inside the hemisphere with radius $R$. I need to use lagrange multipliers. So far I found out that:

$R^2 = c^2 + \frac{a^2}{4} + \frac{b^2}{4}$

$V_{block} = abc$

$V_{hemisphere} = \frac{4}{6}\pi R^3 = \frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2}$

The equation for $R$ is taken from this pictureenter image description here

From here

$L = abc + \lambda\frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2}$

And now I have folowing equations:

$L_a = bc + \lambda\pi \frac{a}{2}(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$

$L_b = ac + \lambda\pi \frac{b}{2}(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$

$L_c = ac + \lambda\pi 2c(c^2 + \frac{a^2}{2} + \frac{b^2}{2})^{1/2} = 0$

$\frac{4}{6}\pi(c^2 + \frac{a^2}{4} + \frac{b^2}{4})^{3/2} = 0$

Which seems unsolveable for me. Did I make mistake somewhere?

Thx for help.

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    $\begingroup$ It is implicit, but you should say it explicitly, that the block has a face on the equatorial plane delimitating the hemisphere. $\endgroup$
    – Jean Marie
    Jun 22, 2017 at 17:21

2 Answers 2

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Your objective functio is $V=abc$ and the constrain is $ R^2=\frac{a^2}{4} + \frac{b^2}{4} + c^2$ then your expression should be

$$L=abc+\lambda\left(\frac{a^2}{4} + \frac{b^2}{4} + c^2 -R^2\right)$$

and then

$$L_a=bc-\lambda\frac{a}{2}=0\to \lambda a=2bc$$

$$L_b=ac-\lambda\frac{b}{2}=0\to \lambda b=2ac$$

$$L_c=ab-2\lambda c=0\to \lambda c=\frac{ab}{2}$$

Can you finish?

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  • $\begingroup$ I can, thank you very much. $\endgroup$
    – Bobesh
    Jun 22, 2017 at 20:29
  • $\begingroup$ You are very welcome $\endgroup$
    – Arnaldo
    Jun 22, 2017 at 21:21
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The formula for $V_{\textrm{hemisphere}}$ is unnecessary---this information is already encoded in the radius $R$ that appears in the constraint. In any case you're aiming to maximize the quantity $$V = a b c$$ subject to the condition $$R^2 = \frac{a^2}{4} + \frac{b^2}{4} + c^2 .$$ Note that by parameterizing our blocks in a slightly different way, namely with $a = 2 A$, $b = 2 B$, $c = C$, we get the similar problem of trying to maximize the quantity $$V = 4 A B C$$ subject to the condition $$R^2 = A^2 + B^2 + C^2 .$$ This formulation has the advantage of being symmetric in $A, B, C$.

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  • $\begingroup$ Great answer, thank you. $\endgroup$
    – Bobesh
    Jun 22, 2017 at 20:29
  • $\begingroup$ You're welcome. $\endgroup$ Jun 22, 2017 at 21:04

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