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3 years ago

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My base is 10. My exponent is an odd number. If you write a multiplication sentence to show my value, it will have fewer than 4

factors of 10. My value is greater than 10. What number am I?

1 answer:

svet-max [94.6K]3 years ago

4 0

. The number is either 10^1 or 10^3 or 10^ 5 or . . . .

<span>If it is more than 10, then it is not 10^1. The next possibility is 10^3which is 1000, which has factors 2, 5, 10, 20, 25, 40, 50, 80, . . . . . </span>

<span>What constitutes a "multiplication sentence" exactly ? This is not a standard mathematical term. Do you mean "reduce it to its prime factors" ? </span>

<span>1000 = 2 x 2 x 2 x 5 x 5 x 5 </span>

<span>which is a lot more than 4 factors. Or not, if you write it as 2³ x 5³. </span>

<span>But you could say the same about 100 000 = 2^5 x 5^5 </span>

<span>or about 10 000 000 = 2^7 x 5^7 </span>

<span>Take your pick. This is a very badly stated question. </span>

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Given surface area of a right circular cone of radious r and height h is,

$S=\pi r\sqrt{r^2+h^2}$

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To find the largest volume if the surface area is S=8 (say), then applying Lagranges multipliers,

$f(r,h)=\frac{1}{3}\pi r^2 h$

subject to,

$g(r,h)=\pi r\sqrt{r^2+h^2}=8\hfill (1)$

We know for maximum volume $r\neq 0$ . So let $\lambda$ be the Lagranges multipliers be such that,

$f_r=\lambda g_r$

$\implies \frac{2}{3}\pi r h=\lambda (\pi \sqrt{r^2+h^2}+\frac{\pi r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}r h= \lambda (\sqrt{r^2+h^2}+\frac{ r^2}{\sqrt{r^2+h^2}})\hfill (2)$

And,

$f_h=\lambda g_h$

$\implies \frac{1}{3}\pi r^2=\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}$

$\implies \lambda=\frac{r\sqrt{r^2+h^2}}{3h}\hfill (3)$

Substitute (3) in (2) we get,

$\frac{2}{3}rh=\frac{r\sqrt{R^2+h^2}}{3h}(\sqrt{R^2+h^2+}+\frac{r^2}{\sqrt{r^2+h^2}})$

$\implies \frac{2}{3}rh=\frac{r}{3h}(2r^2+h^2)$

$\implies h^2=2r^2$

Substitute this value in (1) we get,

$\pi r\sqrt{h^2+r^2}=8$

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Then,

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Hence largest volume,

$V=\frac{1}{3}\times \pi \times\frac{\pi}{8\sqrt{3}}\times 1.71476=0.407114$

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