calculus problem: minimizing cost of tank

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diedead

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A tank with a rectangular base and rectangular sides is to be open at the top. Itis to be constructed so that its width is 4 meters and volume is 36 cubic meters. Ifbuilding the tank costs $10 per square meter for the base and $5 per squaremeter for the sides, what is the cost of the least expensive tank?

i got the answer as 150*root2
i dont know if it's right
please someone check if it's right

what i did was:
V=36=4xz
C=40x+20z
C=360/z + 20z
C'=(-360/x^2) +20
X=3root2
Z=(3root2)/2
so the cost is...C=120root2 + 30root2 = 150root2

thanks in advance
 
let the width be W
let the height be H

cost of base = 4w[10]
cost of a side is 4h[5]
cost of other sides are wh[5]

total cost = 4w[10] +2[4h][5]+2[wh5]
total cost = 40 w +40h+10wh

but volume,[4hw]=36
or h=9/W substituting yields
total cost = 40W+40[9/W]+10w[9/W]
total cost = 40W +360W^-1+90 take derivative
der[total cost] = 40 -360W^-2 set equal to 0
360/W^2=40
W^2=9
W=3 and fro h=9/W
h=3

total cost is 10[12] +2[12][5]+2[9][5]
total cost = 120 +120+90
total cost =$ 330

please check for errors
Arthur
 
Hello, diedead!

Your cost function is incorrect . . .

A tank with a rectangular base and rectangular sides is to be open at the top.
It is to be constructed so that its width is 4 meters and volume is 36 cubic meters.
If building the tank costs $10 per m² for the base and $5 per m² for the sides,
what is the cost of the least expensive tank?
Click to expand...
Code: 
          *-----------*
        / |         / |
      *-----------*   |z
      |           |   |
      |           |   *
      |           | /4
      *-----------*
            x

The bottom has an area of \(\displaystyle 4x\) m².
At $10 per m², it cost is: \(\displaystyle \,40x\) dollars.

The front and back have a total area of \(\displaystyle 2(xz)\) m².
At $5 per m², the cost is: \(\displaystyle 10xz\) dollars.

The left and right have a total area of \(\displaystyle 2(4z)\) m².
At $5 per m², the cost is: \(\displaystyle 40z\) dollars.

Hence, the total cost is: \(\displaystyle \,C\;=\;40x\,+\,10xz\,+\,40z\;\) [1]


The volume is: \(\displaystyle \,4xz\:=\:36\;\;\Rightarrow\;\;x\:=\:\frac{9}{z}\;\) [2]

Substitute [2] into [1]: \(\displaystyle \:C\;=\;40\left(\frac{9}{z}\right)\,+\,10\left(\frac{9}{z}\right)z\,+\,40z\)

. . and we have: \(\displaystyle \:C\;=\;360z^{-1}\,+\,90\,+\,40z\)

Differentiate and equate to zero: \(\displaystyle \:C'\;=\;-360z^{-2}\,+\,40\;=\;0\)

Multiply by \(\displaystyle z^2:\;\;-360\,+\,40z^2\:=\:0\;\;\Rightarrow\;\;z^2\:=\:9\quad\Rightarrow\quad z\,=\,\pm3\)


Discarding the negative answer: \(\displaystyle \,\fbox{z\,=\,3}\)
Substitute into [2]: \(\displaystyle \,\fbox{x\:=\:3}\)


Hence, the dimensions of the box are: \(\displaystyle \,L\,\times\,W\,\times\,H\:=\:3\,\times\,4\,\times\,3\)

The minimum cost is: \(\displaystyle \,C\:=\:\frac{360}{3}\,+\,90\,+\,40(3)\:=\:\L\fbox{\$330}\)


Edit: Too fast for me, Arthur!
 
I agree that the size is 3x3x4, and the cost is $330
If I made a error in calculating the cost , I am sorry and should have checked my results. I do know I calculated the size correctly according to my scrap paper.
Arthur
 
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