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
[tex]V = Length \times Width \times Height \\\\[/tex]
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
[tex]Width = 3 - 2x \\\\[/tex]
The length of the shaded region is given by
[tex]Length = \frac{1}{2} (5 - 3x) \\\\[/tex]
So, the volume of the box becomes,
[tex]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) \\\\[/tex]
In order to maximize the volume enclosed by the box, take the derivative of volume and set it to zero.
[tex]\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 \\\\[/tex]
We are left with a quadratic equation.
We may solve the quadratic equation using quadratic formula.
The quadratic formula is given by
[tex]$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$[/tex]
Where
[tex]a = 18 \\\\b = -38 \\\\c = 15 \\\\[/tex]
[tex]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 \\\\[/tex]
Volume of the box at x= 1.59:
[tex]V = \frac{1}{2} (5 – 3(1.59)) \times (3 - 2(1.59)) \times (1.59) \\\\V = -0.03 \: cm^3 \\\\[/tex]
Volume of the box at x= 0.53:
[tex]V = \frac{1}{2} (5 – 3(0.53)) \times (3 - 2(0.53)) \times (0.53) \\\\V = 1.75 \: cm^3[/tex]
The volume of the box is maximized when x = 0.53 cm
Therefore,
x = 0.53 cm
Maximum volume = 1.75 cm³
