5 times 5 another easy one

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

kirill115 [55]

Answer:

Required largest volume is 0.407114 unit.

Step-by-step explanation:

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

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

and volume,

$V=\frac{1}{3}\pi r^2 h$

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,

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

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

$\implies r=\sqrt{\frac{8}{\pi\sqrt{3}}}\equiv 1.21252$

Then,

$h=\sqrt{2}(1.21252)\equiv 1.71476$

Hence largest volume,

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

3 0

2 years ago

If the total surface area of a cube is 302.46ft2 what best describes the length of the side of the cube?

12345 [234]

Answer:

7.1ft

Step-by-step explanation:

The formula for calculating the total surface area of a cube is 2(L²+L²+L²)

=2(3L²)

= 6L²

Where L is the length of the cube

If the total surface area of a cube is 302.46ft², then

302.46 = 6L²

L² = 302.46/6

L² = 50.41

L =√50.41

L = 7.1ft

The length of the side of the cube is 7.1ft

4 0

3 years ago

HELP QUICK!!!! PLEASE!

Tresset [83]

Answer:

Linear function

Step-by-step explanation:

<h2> $y = \frac{x}{2} - 5$ </h2><h3>Linear function</h3>

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1 and the degree of variable x is 1.

<h2> $y = \frac{2}{x} + 3$ </h2><h3>Not linear function</h3>

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1

, the degrees of the variables in the equation violate the linear equation definition, which means that the equation is not a linear equation.

<h2> $y = {2}^{x} - 1$ </h2><h3>Not linear function</h3>

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degrees of the variables in the equation violate the linear equation definition, which means that the equation is not a linear equation.

<h2> $y = {x}^{2} + 7$ </h2><h3>Not linear function</h3>

A linear equation is an equation of a straight line, which means that the degree of a linear equation must be 0 or 1 for each of its variables. In this case, the degree of variable y is 1 and the degree of variable x is 2.

<h3>Hope it is helpful...</h3>

6 0

2 years ago

Dan buys a car for £1700.

aleksley [76]

Answer:

£1443.89

Step-by-step explanation:

To start you take the £1700 and multiply it by 4% (or .04) to find how much it depreciates for the first year. For the first year the depreciation £68 so the next year it will be worth £1632 ( £1700 - 68). You do the same thing for the second year but you start with the amount its worth now (£1632) and multiply again by the 4%. The depreciation for the second year is 65.28. Now you take what it was worth at the start of the year (£1632) and subtract the depreciation for the second year (65.28) to get £1566.72. You do the same process again for the third year to end up with a value of £1504.05. Now for the 4th year you will take the value of £1504.05 and again multiply by the depreciation rate of 4% to find the last amount of depreciation which is £60.16. Take your starting value for year 4 (£1504.05) and subtract the amount of depreciation (£60.16) to get your answer of £1443.89.

4 0

3 years ago

If f(x) = 7 + 4x and g(x)= 7x, what is the value of (f/g)(5)

andrew-mc [135]

The value of $(\frac{f}{g})(5)$ is $\frac{27}{35}$

Explanation:

Given that $f(x)=7+4x$ and $g(x)=7x$

<u>The value of </u> $(\frac{f}{g})(5)$ <u>:</u>

The value of $(\frac{f}{g})(5)$ can be determined by substituting x = 5 in the functions f(x) and g(x) and then dividing the terms, we get, the value of $(\frac{f}{g})(5)$

Thus, we have,

$(\frac{f}{g})(5)=\frac{f(5)}{g(5)}$