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Allisa [31]
2 years ago
10

Having trouble figuring out this problem. What function should be used to maximize the volume of an open box whose surface area

is equal to 20 where the box has a square bottom?

Mathematics
1 answer:
omeli [17]2 years ago
4 0

The given constraints of the value of the surface area

determines the values of <em>x</em> and <em>h</em> at the maximum volume.

Correct response:

  • The function to be used to maximize the volume of the open box is; \underline{Volume = x^2 \cdot \dfrac{\sqrt{10} }{4}}

  • At the maximum volume, \underline{x = \sqrt{10} }, \underline{h = \dfrac{\sqrt{10} }{4} }

<h3>Which is the method used to find the maximum volume of the box</h3>

The given parameters are;

Length of the side of the square base of the box = x

Height of the box = h

Surface area of the box = 20

Required:

Function that can be used to maximize the volume.

Solution:

The function for the surface area is; S.A. = 4·h·x + x²

Volume of the box, V = x²·h

The given surface area, S.A. = 20

Therefore;

S.A. = 20 = 4·h·x + x²

  • h = \mathbf{\dfrac{20 - x^2}{4 \cdot x}}

Which gives;

The volume as a function of <em>x</em> is given as follows;

V = x^2 \times \left(\dfrac{20 - x^2}{4 \cdot x} \right) = \mathbf{5\cdot x -  \dfrac{x^3}{4}}

At the maximum volume, we have;

\dfrac{dV}{dx} = 0 = \dfrac{d}{dx} \left(5 \cdot x - \dfrac{x^3}{4} \right) = 5 - 2 \cdot \dfrac{x^2}{4}  = \mathbf{ 5 - \dfrac{x^2}{2}}

5 - \dfrac{x^2}{2}  = 0

\dfrac{x^2}{2} = 5

x² = 10

x = √10

Second \ derivative, \ f''(x) = \dfrac{d}{dx} \left(5 - \dfrac{x^2}{2} \right) = \mathbf{-x}

f''(√10) = -√10 < 0, therefore;

The maximum value of the volume is given at <u><em>x</em></u><u> = √10</u>

Therefore;

At \ maximum \ volume, \ h = \mathbf{\dfrac{20 - \left(\sqrt{10}^2 \right)}{4 \times \sqrt{10} }}  = \dfrac{10}{4 \cdot \sqrt{10} }  = \dfrac{\sqrt{10} }{4}

At the maximum x = √(10), h = \dfrac{\sqrt{10} }{4}

  • A \ function \ for \ maximum \ volume \ is, \ V = \underline{x^2 \cdot \dfrac{\sqrt{10} }{4} }

The values of <em>x</em> and <em>h</em> at the maximum volume are;

  • \underline{x = \sqrt{10} } and \underline{h = \dfrac{\sqrt{10} }{4} }

Learn more about maximum value of a function here:

brainly.com/question/10782019

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