Answer:
Width is 3 ft
length is 6 ft
Step-by-step explanation:
Let x = the width
Let 2x = the length
Let h = the height
We know that volume is given v =  l×b×h
so we have v = volume  [tex]= x*2x*h[/tex]
[tex]2x^2*h = 36[/tex]
[tex]h = \frac{18}{x^2}[/tex]
surface area include 2 ends + 1 bottom and 2 sides excluding top
[tex]S.A. = 2(x*h) + 1(2x*x) + 2(2x*h)[/tex]
[tex]S.A. = 2xh + 2x^2 + 4xh[/tex]
we have calculate[tex] h = = \frac{18}{x^2}[/tex]
[tex]S.A = 2x^2 + 6x \frac{18}{x^2}
[/tex]
[tex]S.A = 2x^2 + \frac{108}{x}[/tex]
By graphing above equation we can calculate the value of x
Minimum S.A when x is 3 Â is the width so
2(3) = 6 is the length
[tex]h = \frac{18}{3^2} = 2[/tex]
The volume of the box is the amount of material it can contain.
The width of the box that can be produced using the minimum amount of material is 3 ft.
The volume of the box is give as:
[tex]\mathbf{V = 36}[/tex]
The dimension of the cardboard is
[tex]\mathbf{l= 2w}[/tex]
The volume is calculated as:
[tex]\mathbf{V = lwh}[/tex]
Substitute [tex]\mathbf{l= 2w}[/tex]
[tex]\mathbf{V = 2w^2h}[/tex]
Substitute [tex]\mathbf{V = 36}[/tex]
[tex]\mathbf{2w^2h = 36}[/tex]
Divide through by 2
[tex]\mathbf{w^2h = 18}[/tex]
Make h the subject
[tex]\mathbf{h = \frac{18}{w^2} }[/tex]
The surface area is calculated as:
[tex]\mathbf{S = (2wh + lw + 2lh)}[/tex]
Substitute [tex]\mathbf{h = \frac{18}{w^2} }[/tex]
[tex]\mathbf{S = (2w\times \frac{18}{w^2} + lw + 2l\times \frac{18}{w^2})}[/tex]
[tex]\mathbf{S = \frac{36}{w} + lw + \frac{36l}{w^2}}[/tex]
Substitute [tex]\mathbf{l= 2w}[/tex]
[tex]\mathbf{S = \frac{36}{w} + 2w^2 + \frac{72w}{w^2}}[/tex]
[tex]\mathbf{S = \frac{36}{w} + 2w^2 + \frac{72}{w}}[/tex]
[tex]\mathbf{S = 2w^2 + \frac{108}{w}}[/tex]
Differentiate
[tex]\mathbf{S' = 4w - \frac{108}{w^2}}[/tex]
Set to 0
[tex]\mathbf{4w = \frac{108}{w^2}}[/tex]
Multiply through by w^2
[tex]\mathbf{4w^3 = 108}[/tex]
Divide through by 4
[tex]\mathbf{w^3 = 27}[/tex]
Take cube roots
[tex]\mathbf{w = 3}[/tex]
Hence, the width of the box is 3 ft.
Read more about volumes at:
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