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Lesechka [4]
3 years ago
7

Consider the function V=g(x), where g(x) =x(6-2x)(8-2x), with x being the length of a cutout in cm and V being the volume of an

open box in cm³, where x must between 0 and 3 cm. Determine the maximum volume of the open box in cm³. Round your answer to two decimal places. Also, indicate or show what process you used to obtain this answer. (Hint: graphing)

Mathematics
1 answer:
Andrej [43]3 years ago
6 0

Answer:

The maximum volume of the open box is 24.26 cm³

Step-by-step explanation:

The volume of the box is given as V=g(x), where g(x)=x(6-2x)(8-2x) and 0\le x\le3.

Expand the function to obtain:

g(x)=4x^3-28x^2+48x

Differentiate  wrt  x to obtain:

g'(x)=12x^2-56x+48

To find the point where the maximum value occurs, we solve

g'(x)=0

\implies 12x^2-56x+48=0

\implies x=1.13,x=3.54

Discard x=3.54 because it is not within the given domain.

Apply the second derivative test to confirm the maximum critical point.

g''(x)=24x-56, g''(1.13)=24(1.13)-56=-28.88\:

This means the maximum volume occurs at x=1.13.

Substitute x=1.13 into g(x)=x(6-2x)(8-2x) to get the maximum volume.

g(1.13)=1.13(6-2\times1.13)(8-2\times1.13)=24.26

The maximum volume of the open box is 24.26 cm³

See attachment for graph.

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Answer:

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Step-by-step explanation:

As the complete question is not given, thus the complete question is found online and is attached herewith.

So the joint density function is given as

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f(Y/X)=\dfrac{f(X,Y)}{f(X)}\\f(Y/X)=\dfrac{\dfrac{x}{8}e^{-\frac{x+y}{2} }}{\dfrac{x}{4}e^{-\frac{x}{2}}}\\f(Y/X)=\dfrac{1}{2}e^{-\frac{y}{2} }

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So the value of E(Y/X) is 2.

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