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1 answer:

5 0

SOLUTION:

A box has a square base of side x and height y where x,y > 0.
Its volume is V = x^2y and its surface area is S = 2x^2 + 4xy.
(a) If V = x^2y = 12, then y = 12=x^2 and S(x) = 2x^2 C 4x (12=x2) = 2x2 + 48x^-1. Solve S'(x) = 4x - 48x-2 = 0 to obtain x = 12^1/3. Since S(x/) ---> infinite as x ---> 0+ and as x --->infinite, the minimum surface area is S(12^1/3) = 6 (12)^2/3 = 31.45, when x = 12^1/3 and y = 12^1/3.

(b) If S = 2x2 + 4xy = 20, then y = 5^x-1 - 1/2 x and V (x) = x^2y = 5x - 1/2x^3. Note that x must lie on the closed interval [0, square root of 10]. Solve V' (x) = 5 - 3/2 x^2 for x>0 to obtain x = square root of 30 over 3 . Since V(0) = V (square root 10) = 0 and V(square root 30 over 3) = 10 square root 30 over 9 , the maximum volume is V (square root 30 over 3) = 10/9 square root 30 = 6.086, when x = square root 30 over 3 and y = square root 30 over 3 .

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Which integer is closest to 0 on the number line?

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3 years ago

Read 2 more answers

In ΔLMN, \overline{LN}

Karolina [17]

Answer:

$\bold{\angle NLM = 10^\circ}$

Step-by-step explanation:

Given a ΔLMN.

Line LN is extended to point O.

such that:

$\text{m}\angle MNO = (5x-13)^{\circ}$

$\text{m}\angle NLM = (x-4)^{\circ}$ and

$\text{m}\angle LMN = (2x+19)^{\circ}$

To find:

$\text{m}\angle NLM=?$

Solution:

Kindly refer to the attached image for the given triangle and dimensions of angles.

Let us recall the external angle property of a triangle:

The external angle of a triangle is equal to the sum of two opposite internal angles.

i.e.

$\angle MNO = \angle NLM + \angle LMN\\\Rightarrow 5x-13 = x-4+2x+19\\\Rightarrow 5x-13 = 3x+15\\\Rightarrow 5x-3x = 13+15\\\Rightarrow 2x=28\\\Rightarrow x =14$