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}}](https://tex.z-dn.net/?f=%5Cunderline%7BVolume%20%3D%20x%5E2%20%5Ccdot%20%5Cdfrac%7B%5Csqrt%7B10%7D%20%7D%7B4%7D%7D)
- At the maximum volume,
, ![\underline{h = \dfrac{\sqrt{10} }{4} }](https://tex.z-dn.net/?f=%5Cunderline%7Bh%20%3D%20%5Cdfrac%7B%5Csqrt%7B10%7D%20%7D%7B4%7D%20%7D)
<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
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}}](https://tex.z-dn.net/?f=V%20%3D%20x%5E2%20%5Ctimes%20%5Cleft%28%5Cdfrac%7B20%20-%20x%5E2%7D%7B4%20%5Ccdot%20x%7D%20%5Cright%29%20%3D%20%5Cmathbf%7B5%5Ccdot%20x%20-%20%20%5Cdfrac%7Bx%5E3%7D%7B4%7D%7D)
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}}](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdx%7D%20%3D%200%20%3D%20%5Cdfrac%7Bd%7D%7Bdx%7D%20%5Cleft%285%20%5Ccdot%20x%20-%20%5Cdfrac%7Bx%5E3%7D%7B4%7D%20%5Cright%29%20%3D%205%20-%202%20%5Ccdot%20%5Cdfrac%7Bx%5E2%7D%7B4%7D%20%20%3D%20%5Cmathbf%7B%205%20-%20%5Cdfrac%7Bx%5E2%7D%7B2%7D%7D)
![5 - \dfrac{x^2}{2} = 0](https://tex.z-dn.net/?f=5%20-%20%5Cdfrac%7Bx%5E2%7D%7B2%7D%20%20%3D%200)
![\dfrac{x^2}{2} = 5](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%7D%7B2%7D%20%3D%205)
x² = 10
x = √10
![Second \ derivative, \ f''(x) = \dfrac{d}{dx} \left(5 - \dfrac{x^2}{2} \right) = \mathbf{-x}](https://tex.z-dn.net/?f=Second%20%5C%20derivative%2C%20%5C%20f%27%27%28x%29%20%3D%20%5Cdfrac%7Bd%7D%7Bdx%7D%20%5Cleft%285%20-%20%5Cdfrac%7Bx%5E2%7D%7B2%7D%20%5Cright%29%20%3D%20%5Cmathbf%7B-x%7D)
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}](https://tex.z-dn.net/?f=At%20%5C%20maximum%20%5C%20volume%2C%20%5C%20h%20%3D%20%5Cmathbf%7B%5Cdfrac%7B20%20-%20%5Cleft%28%5Csqrt%7B10%7D%5E2%20%5Cright%29%7D%7B4%20%5Ctimes%20%5Csqrt%7B10%7D%20%7D%7D%20%20%3D%20%5Cdfrac%7B10%7D%7B4%20%5Ccdot%20%5Csqrt%7B10%7D%20%7D%20%20%3D%20%5Cdfrac%7B%5Csqrt%7B10%7D%20%7D%7B4%7D)
At the maximum x = √(10), h = ![\dfrac{\sqrt{10} }{4}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%7B10%7D%20%7D%7B4%7D)
The values of <em>x</em> and <em>h</em> at the maximum volume are;
and ![\underline{h = \dfrac{\sqrt{10} }{4} }](https://tex.z-dn.net/?f=%5Cunderline%7Bh%20%3D%20%5Cdfrac%7B%5Csqrt%7B10%7D%20%7D%7B4%7D%20%7D)
Learn more about maximum value of a function here:
brainly.com/question/10782019