A company wishes to manufacture a box with a volume of 2424 cubic feet that is open on top and is twice as long as it is wide. f
ind the width of the box that can be produced using the minimum amount of material
1 answer:
<span> Let x = the width
Let 2x = the length
Let h = the height
</span>then vol = x*2x*h. So we have 2x^2*h = 24
h=24/(2*x^2)=12/x^2
Surface area: two ends + 1 bottom + 2 sides (no top)
S.A. = 2(x*h) + 1(2x*x) + 2(2x*h)
S.A. = 2xh + 2x^2 + 4xh<span> S.A. = 2x^2 + 6xh
</span>Replace h with 12/x^2
S.A = 2x^2 + 6x(12/x^2)
S.A = 2x^2 + 6(12/x)
S.A = 2x^2 + (72/x)
Graph this equation to find the value of x for minimum material
Min surface area when x = 3.0 is the width<span> then
</span>2(3) = 6 is the length
Find the height:
h=12/(3.0)^2
h=1.33
Box dimensions for min surface area: 3.0 by 6 by 1.33; much better numbers
Check the vol of these dimensions: 3.0*6*1.33 ~ 24
graphic attachment
Let 2x = the length
Let h = the height
</span>then vol = x*2x*h. So we have 2x^2*h = 24
h=24/(2*x^2)=12/x^2
Surface area: two ends + 1 bottom + 2 sides (no top)
S.A. = 2(x*h) + 1(2x*x) + 2(2x*h)
S.A. = 2xh + 2x^2 + 4xh<span> S.A. = 2x^2 + 6xh
</span>Replace h with 12/x^2
S.A = 2x^2 + 6x(12/x^2)
S.A = 2x^2 + 6(12/x)
S.A = 2x^2 + (72/x)
Graph this equation to find the value of x for minimum material
Min surface area when x = 3.0 is the width<span> then
</span>2(3) = 6 is the length
Find the height:
h=12/(3.0)^2
h=1.33
Box dimensions for min surface area: 3.0 by 6 by 1.33; much better numbers
Check the vol of these dimensions: 3.0*6*1.33 ~ 24
graphic attachment
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