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3 years ago

HELP ME PLEASE!!! THANK YOU!

Mathematics

1 answer:

AURORKA [14]3 years ago

3 0

Answer:

down v

Step-by-step explanation:

1=20 mph

2=40 mph

3 1/4= 60.5 mph(im not sure if its correct but.)

.5 mph=10

2.5 mph=50

thats all i know

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Determine the dimensions of a rectangular solid (with a square base) with maximum volume if its surface area is 181.5 square cen

3241004551 [841]

Answer:

The base and height of the solid is 5.5cm

Step-by-step explanation:

Given

$Surface\ Area = 181.5cm^2$

Required

Determine the dimensions that maximizes the volume

Let the base dimension be x and the height be h

The volume is calculated as:

$Volume =x^2 * h$

$Volume =x^2h$

181.5 =x^2h

The surface area (S) is calculated as this:

$S = 2(x^2 + xh + xh)$

$S = 2(x^2 + 2xh)$

$S = 2x^2 + 4xh$

Substitute 181.5 for S

$181.5 = 2x^2 + 4xh$

Make h the subject:

$4xh = 181.5 - 2x^2$

$h = \frac{181.5 - 2x^2}{4x}$

Substitute $h = \frac{181.5 - 2x^2}{4x}$ in $Volume =x^2h$

$V = x^2(\frac{181.5 - 2x^2}{4x})$

$V = x(\frac{181.5 - 2x^2}{4})$

$V = \frac{1}{4}(x)(181.5 - 2x^2)$

$V = \frac{181.5x}{4} - \frac{2x^3}{4}$

$V = \frac{181.5x}{4} - \frac{x^3}{2}$

To get the maximum, we differentiate V with respect to t and set the differentiation to 0

$dV = \frac{181.5}{4} - \frac{3x^2}{2}$

Set to 0

$0 = \frac{181.5}{4} - \frac{3x^2}{2}$

$\frac{3x^2}{2} = \frac{181.5}{4}$

Multiply through by 4

$4 * \frac{3x^2}{2} = \frac{181.5}{4} * 4$

$2*3x^2 = 181.5$

$6x^2 = 181.5$

$x^2 = \frac{181.5}{6}$

$x^2 = 30.25$

$x = \sqrt{30.25$

$x = 5.5$

Recall that:

$h = \frac{181.5 - 2x^2}{4x}$

$h = \frac{181.5 - 2 * 5.5^2}{4 * 5.5}$

$h = \frac{181.5 - 60.5}{22}$

$h = \frac{121}{22}$

$h = 5.5$

So, we have:

$h = 5.5$

$x = 5.5$

<em>Hence, the base and height of the solid is 5.5cm</em>

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Why?
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