What steps would you use to determine the dimensions that will minimize the cost to build a box using rationals? The question wants us to find the minimal cost to make a box whose base length is the same as the base width. The material used for the top and bottom cats $2/cm2 and the material for the sides cats $1/cm2. The box has to have a volume of 30cm3.
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calendarEducator since 2011
write562 answers
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w = width
h = height
The length is equal to the width
Area of top and bottom = 2w^2
Area of sides = 2wh + 2wh = 4wh
So
Total cost `= 2w^2*2 + (4wh)1 = 4w^2 + 4wh`
The total volume is `30cm^3 = w^2h`
So `h=30/(w^2)` substituting we get
Total cost `= 2w^2 + 4w*30/w^2 = 2w^2 + 120/w`
`(d(Total cos))/(dw) = 4w - 120/w^2`
`(d(Total cost))/(dw) = 0` when
`4w - 120/w^2 = 0`
` 4w = 120/w^2`
`w^3 = 30`
`w = root(3)(30)`
`h = 30/w^2 = 30/(root(3)(30))^2 = root(3)(30)`
I am not sure what your comment about rationals. You can estimate these numbers to any number of decimals. That would be rational answers. The answer is w = 3.107 cm, h = 3.107 cm.
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calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
The dimensions of the box are : L, W, and H
We know that L= W.
==> V = L*H*W = L^2 H = 30 ==>
==> H = 30/L^2
==> Surface area for the top and bottom = L*W*2 = 2L^2
==> Cost for top and bottom = 2L^2 * 2 = 4L^2
==> Surface area for the sides = 2HW + 2HL = 2HL+2HL = 4HL
==> Cost for sides = 4HL*1 = 4HL
But H= 30/L^2
==> cost for sides = 4L *30/L^2 = 120/L
==> Total cost = 2L^2 + 120/L
==> f(L) = 2L^2 + 120/L
Now we will find the minimum points:
We will find the derivative;s zeros which given the extreme value.
==> f'(L)= 4L -120/L^2 = 0
==> 4L = 120/L^2
==> 4L^3 = 120
==> L^3 = 30
==> L= 3.12
==> H= 30/30= 1
==> W = L = 3.12
==> The dimensions are: 3.12, 3.12 , and 1.