Answer:
The base and height of the solid is 5.5cm
Step-by-step explanation:
Given
[tex]Surface\ Area = 181.5cm^2[/tex]
Required
Determine the dimensions that maximizes the volume
Let the base dimension be x and the height be h
The volume is calculated as:
[tex]Volume =x^2 * h[/tex]
[tex]Volume =x^2h[/tex]
181.5 =x^2h
The surface area (S) is calculated as this:
[tex]S = 2(x^2 + xh + xh)[/tex]
[tex]S = 2(x^2 + 2xh)[/tex]
[tex]S = 2x^2 + 4xh[/tex]
Substitute 181.5 for S
[tex]181.5 = 2x^2 + 4xh[/tex]
Make h the subject:
[tex]4xh = 181.5 - 2x^2[/tex]
[tex]h = \frac{181.5 - 2x^2}{4x}[/tex]
Substitute [tex]h = \frac{181.5 - 2x^2}{4x}[/tex] in [tex]Volume =x^2h[/tex]
[tex]V = x^2(\frac{181.5 - 2x^2}{4x})[/tex]
[tex]V = x(\frac{181.5 - 2x^2}{4})[/tex]
[tex]V = \frac{1}{4}(x)(181.5 - 2x^2)[/tex]
[tex]V = \frac{181.5x}{4} - \frac{2x^3}{4}[/tex]
[tex]V = \frac{181.5x}{4} - \frac{x^3}{2}[/tex]
To get the maximum, we differentiate V with respect to t and set the differentiation to 0
[tex]dV = \frac{181.5}{4} - \frac{3x^2}{2}[/tex]
Set to 0
[tex]0 = \frac{181.5}{4} - \frac{3x^2}{2}[/tex]
[tex]\frac{3x^2}{2} = \frac{181.5}{4}[/tex]
Multiply through by 4
[tex]4 * \frac{3x^2}{2} = \frac{181.5}{4} * 4[/tex]
[tex]2*3x^2 = 181.5[/tex]
[tex]6x^2 = 181.5[/tex]
[tex]x^2 = \frac{181.5}{6}[/tex]
[tex]x^2 = 30.25[/tex]
[tex]x = \sqrt{30.25[/tex]
[tex]x = 5.5[/tex]
Recall that:
[tex]h = \frac{181.5 - 2x^2}{4x}[/tex]
[tex]h = \frac{181.5 - 2 * 5.5^2}{4 * 5.5}[/tex]
[tex]h = \frac{181.5 - 60.5}{22}[/tex]
[tex]h = \frac{121}{22}[/tex]
[tex]h = 5.5[/tex]
So, we have:
[tex]h = 5.5[/tex]
[tex]x = 5.5[/tex]
Hence, the base and height of the solid is 5.5cm
