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tankabanditka [31]

3 years ago

8

Which set of ordered pairs does NOT represent a function? A) (0, 1), (2, 2), (4, 8), (-2, 7), (5, 8) B) (0, 1), (2, 2), (4, 8),

(2, 7), (5, 8), (7, 9) C) (-3, 6), (2, 7), (0, 5), (1, 5), (4, 9), (5, 4) D) (-4, 2), (-3, 2), (-2, 2), (-1, 2), (0, 2), (1, 2) Submit

1 answer:

Viktor [21]3 years ago

8 0

That is B
Because there are duplicate x-values in the ordered pairs, ( the (2,2) and the (2,7).

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The surface area of a right circular cone of radius r and height h is S = πr√ r 2 + h 2 , and its volume is V = 1 3 πr2h. What i

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

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

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

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

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

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

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,

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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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