The given constraints of the value of the surface area
determines the values of x and h at the maximum volume.
Correct response:
- The function to be used to maximize the volume of the open box is; [tex]\underline{Volume = x^2 \cdot \dfrac{\sqrt{10} }{4}}[/tex]
- At the maximum volume, [tex]\underline{x = \sqrt{10} }[/tex], [tex]\underline{h = \dfrac{\sqrt{10} }{4} }[/tex]
Which is the method used to find the maximum volume of the box
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
- [tex]h = \mathbf{\dfrac{20 - x^2}{4 \cdot x}}[/tex]
Which gives;
The volume as a function of x is given as follows;
[tex]V = x^2 \times \left(\dfrac{20 - x^2}{4 \cdot x} \right) = \mathbf{5\cdot x - \dfrac{x^3}{4}}[/tex]
At the maximum volume, we have;
[tex]\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}}[/tex]
[tex]5 - \dfrac{x^2}{2} = 0[/tex]
[tex]\dfrac{x^2}{2} = 5[/tex]
x² = 10
x = √10
[tex]Second \ derivative, \ f''(x) = \dfrac{d}{dx} \left(5 - \dfrac{x^2}{2} \right) = \mathbf{-x}[/tex]
f''(√10) = -√10 < 0, therefore;
The maximum value of the volume is given at x = √10
Therefore;
[tex]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}[/tex]
At the maximum x = √(10), h = [tex]\dfrac{\sqrt{10} }{4}[/tex]
- [tex]A \ function \ for \ maximum \ volume \ is, \ V = \underline{x^2 \cdot \dfrac{\sqrt{10} }{4} }[/tex]
The values of x and h at the maximum volume are;
- [tex]\underline{x = \sqrt{10} }[/tex] and [tex]\underline{h = \dfrac{\sqrt{10} }{4} }[/tex]
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