Answer:
x = 0.53 cm
Maximum volume = 1.75 cm³
Step-by-step explanation:
Refer to the attached diagram:
The volume of the box is given by
![V = Length \times Width \times Height \\\\](https://tex.z-dn.net/?f=V%20%3D%20Length%20%5Ctimes%20Width%20%5Ctimes%20Height%20%5C%5C%5C%5C)
Let x denote the length of the sides of the square as shown in the diagram.
The width of the shaded region is given by
![Width = 3 - 2x \\\\](https://tex.z-dn.net/?f=Width%20%3D%203%20-%202x%20%5C%5C%5C%5C)
The length of the shaded region is given by
![Length = \frac{1}{2} (5 - 3x) \\\\](https://tex.z-dn.net/?f=Length%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%285%20-%203x%29%20%5C%5C%5C%5C)
So, the volume of the box becomes,
![V = \frac{1}{2} (5 - 3x) \times (3 - 2x) \times x \\\\V = \frac{1}{2} (5 - 3x) \times (3x - 2x^2) \\\\V = \frac{1}{2} (15x -10x^2 -9 x^2 + 6 x^3) \\\\V = \frac{1}{2} (6x^3 -19x^2 + 15x) \\\\](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%285%20-%203x%29%20%5Ctimes%20%283%20-%202x%29%20%5Ctimes%20x%20%5C%5C%5C%5CV%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%285%20-%203x%29%20%5Ctimes%20%283x%20-%202x%5E2%29%20%5C%5C%5C%5CV%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%2815x%20-10x%5E2%20-9%20x%5E2%20%2B%206%20x%5E3%29%20%5C%5C%5C%5CV%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%286x%5E3%20-19x%5E2%20%2B%2015x%29%20%5C%5C%5C%5C)
In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.
![\frac{dV}{dx} = 0 \\\\\frac{dV}{dx} = \frac{d}{dx} ( \frac{1}{2} (6x^3 -19x^2 + 15x)) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\\frac{dV}{dx} = \frac{1}{2} (18x^2 -38x + 15) \\\\0 = \frac{1}{2} (18x^2 -38x + 15) \\\\18x^2 -38x + 15 = 0 \\\\](https://tex.z-dn.net/?f=%5Cfrac%7BdV%7D%7Bdx%7D%20%3D%200%20%5C%5C%5C%5C%5Cfrac%7BdV%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdx%7D%20%28%20%5Cfrac%7B1%7D%7B2%7D%20%286x%5E3%20-19x%5E2%20%2B%2015x%29%29%20%5C%5C%5C%5C%5Cfrac%7BdV%7D%7Bdx%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2818x%5E2%20-38x%20%2B%2015%29%20%5C%5C%5C%5C%5Cfrac%7BdV%7D%7Bdx%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2818x%5E2%20-38x%20%2B%2015%29%20%5C%5C%5C%5C0%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2818x%5E2%20-38x%20%2B%2015%29%20%5C%5C%5C%5C18x%5E2%20-38x%20%2B%2015%20%3D%200%20%5C%5C%5C%5C)
We are left with a quadratic equation.
We may solve the quadratic equation using quadratic formula.
The quadratic formula is given by
![$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$](https://tex.z-dn.net/?f=%24x%3D%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%24)
Where
![a = 18 \\\\b = -38 \\\\c = 15 \\\\](https://tex.z-dn.net/?f=a%20%3D%2018%20%5C%5C%5C%5Cb%20%3D%20-38%20%5C%5C%5C%5Cc%20%3D%2015%20%5C%5C%5C%5C)
![x=\frac{-(-38)\pm\sqrt{(-38)^2-4(18)(15)}}{2(18)} \\\\x=\frac{38\pm\sqrt{(1444- 1080}}{36} \\\\x=\frac{38\pm\sqrt{(364}}{36} \\\\x=\frac{38\pm 19.078}{36} \\\\x=\frac{38 + 19.078}{36} \: or \: x=\frac{38 - 19.078}{36}\\\\x= 1.59 \: or \: x = 0.53 \\\\](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-%28-38%29%5Cpm%5Csqrt%7B%28-38%29%5E2-4%2818%29%2815%29%7D%7D%7B2%2818%29%7D%20%5C%5C%5C%5Cx%3D%5Cfrac%7B38%5Cpm%5Csqrt%7B%281444-%201080%7D%7D%7B36%7D%20%5C%5C%5C%5Cx%3D%5Cfrac%7B38%5Cpm%5Csqrt%7B%28364%7D%7D%7B36%7D%20%5C%5C%5C%5Cx%3D%5Cfrac%7B38%5Cpm%2019.078%7D%7B36%7D%20%5C%5C%5C%5Cx%3D%5Cfrac%7B38%20%2B%20%2019.078%7D%7B36%7D%20%5C%3A%20or%20%5C%3A%20x%3D%5Cfrac%7B38%20-%2019.078%7D%7B36%7D%5C%5C%5C%5Cx%3D%201.59%20%5C%3A%20or%20%5C%3A%20x%20%3D%200.53%20%5C%5C%5C%5C)
Volume of the box at x= 1.59:
![V = \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%285%20%E2%80%93%203%281.59%29%29%20%5Ctimes%20%283%20-%202%281.59%29%29%20%5Ctimes%20%281.59%29%20%5C%5C%5C%5CV%20%3D%20-0.03%20%5C%3A%20cm%5E3%20%5C%5C%5C%5C)
Volume of the box at x= 0.53:
![V = \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%285%20%E2%80%93%203%280.53%29%29%20%5Ctimes%20%283%20-%202%280.53%29%29%20%5Ctimes%20%280.53%29%20%5C%5C%5C%5CV%20%3D%201.75%20%5C%3A%20cm%5E3)
The volume of the box is maximized when x = 0.53 cm
Therefore,
x = 0.53 cm
Maximum volume = 1.75 cm³