Answer:
[tex]h = 3.5[/tex]
[tex]w = 7[/tex]
[tex]l = 7[/tex]
Step-by-step explanation:
Given
[tex]Volume = 171.5ft^3[/tex]
Required
The dimension that requires least material
The volume is:
[tex]Volume = lwh[/tex]
Where:
[tex]l \to length[/tex]
[tex]w \to width[/tex]
[tex]h \to height[/tex]
So, we have:
[tex]171.5 = lwh[/tex]
Make l the subject
[tex]l = \frac{171.5}{wh}[/tex]
The surface area (A) of an open-top rectangular tank is:
[tex]A = lw + 2lh + 2wh[/tex]
Substitute: [tex]l = \frac{171.5}{wh}[/tex]
[tex]A = \frac{171.5}{wh} * w + 2*\frac{171.5}{wh}*h + 2wh[/tex]
[tex]A = \frac{171.5}{h} + 2*\frac{171.5}{w} + 2wh[/tex]
[tex]A = \frac{171.5}{h} + \frac{343}{w} + 2wh[/tex]
Rewrite as:
[tex]A = 171.5h^{-1} + 343w^{-1} + 2wh[/tex]
Differentiate with respect to h and w
[tex]A_h = -171.5h^{-2} +2w[/tex]
[tex]A_w = -343w^{-2} +2h[/tex]
Equate both to 0
[tex]-171.5h^{-2} +2w=0[/tex]
Make w the subject
[tex]2w = 171.5h^{-2}[/tex]
Divide by 2
[tex]w = 85.75h^{-2}[/tex]
[tex]-343w^{-2} +2h = 0[/tex]
Make h the subject
[tex]2h = 343w^{-2}[/tex]
Divide by 2
[tex]h = 171.5w^{-2}[/tex]
[tex]h = \frac{171.5}{w^2}[/tex]
Substitute [tex]w = 85.75h^{-2}[/tex] in [tex]h = \frac{171.5}{w^2}[/tex]
[tex]h = \frac{171.5}{(85.75h^{-2})^2}[/tex]
[tex]h = \frac{171.5}{85.75^2*h^{-4}}[/tex]
[tex]h = \frac{2}{85.75*h^{-4}}[/tex]
Multiply both sides by [tex]h^{-4}[/tex]
[tex]h^{-4} * h = \frac{1}{85.75*h^{-4}} * h^{-4}[/tex]
[tex]h^{-3} = \frac{2}{85.75}[/tex]
Rewrite as:
[tex]\frac{1}{h^3} = \frac{2}{85.75}[/tex]
Inverse both sides
[tex]h^3 = 85.75/2[/tex]
[tex]h^3 = 42.875[/tex]
Take cube roots
[tex]h = 3.5[/tex] ---- height
Recall that: [tex]w = 85.75h^{-2}[/tex]
[tex]w = 85.75 * 3.5^{-2}[/tex]
[tex]w = 7[/tex] --- width
Recall that: [tex]l = \frac{171.5}{wh}[/tex]
[tex]l = \frac{171.5}{3.5 * 7}[/tex]
[tex]l = 7[/tex] --- length
