an aquarium has a square base made of slate costing 8 cents/in squared and 4 glass sides costing 3 cents/in squared. the volume of the aquarium is to be 36000 inches cubed. find the dimensions of the least expensive such aquarium.

Let x be a side of the square base and y the height. Then we have that

Volume = (x^2)* y
So 36000 =(x^2)* y

If we solve for y we get y =36000/(x^2)

Next we create a function C(x) to represent the total cost

C(x) =8*x^2 + 3*4*xy
=8*x^2 + 12 xy
=8*x^2 + 12*(36000/x) Using the equation for y above
= 8*x^2 + 432000/x
Now to minimize the cost we find C’ (x)

C’(x) = 16x – 432000/x^2

Next Let C’(x) = 0 to find the critical points and we get

16x = 432000/x^2

x^3= 27000 so then x = 30

then y = 36000 / (30^2) =40

So the dimensions should be x=30 and y = 40

One Response

1. Trini Says:
   February 24th, 2010 at 4:51 am

   Let x be a side of the square base and y the height. Then we have that

   Volume = (x^2)* y
   So 36000 =(x^2)* y

   If we solve for y we get y =36000/(x^2)

   Next we create a function C(x) to represent the total cost

   C(x) =8*x^2 + 3*4*xy
   =8*x^2 + 12 xy
   =8*x^2 + 12*(36000/x) Using the equation for y above
   = 8*x^2 + 432000/x
   Now to minimize the cost we find C’ (x)

   C’(x) = 16x – 432000/x^2

   Next Let C’(x) = 0 to find the critical points and we get

   16x = 432000/x^2

   x^3= 27000 so then x = 30

   then y = 36000 / (30^2) =40

   So the dimensions should be x=30 and y = 40
   References :

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