Calculus question - minimize cost from volume

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hello,
im only having one problem with this question..and that is finding my second formula.
I have an aquarium with givin volume, where the base is made of slate and the sides are made of glass..and the cost of slate is 5 times the cost of glass (per unit area) find the dimension of the aquarium to minimize cost.
ok so equation one is
V = lwh l - length w - width h - height
and my second is somthing like...
5(2lh + 2wh) = lw
the first part is the sides of the aquarium and the second part is the base.. now this is comparing the costs of the areas of the sides and the area of the base.. but i don't think it is right because how can i say that the cost of the base is the same of the cost of all 4 sides. seaing how there is more area around the sides then the base...
thanks
 
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Pearce_09 said:
hello,
im only having one problem with this question..and that is finding my second formula.
I have an aquarium with givin volume, where the base is made of slate and the sides are made of glass..and the cost of slate is 5 times the cost of glass (per unit area) find the dimension of the aquarium to minimize cost.
ok so equation one is
V = lwh l - length w - width h - height
and my second is somthing like...
5(2lh + 2wh) = lw
NO. The total cost of the aquarium will be the cost of the 4 sides:
2lh+ 2wh
and the cost of the base: 5wh.

the first part is the sides of the aquarium and the second part is the base.. now this is comparing the costs of the areas of the sides and the area of the base..
But you multiplied the all by 5. How does that take into account that the base costs 5 times a much as the sides?

but i don't think it is right because how can i say that the cost of the base is the same of the cost of all 4 sides. seaing how there is more area around the sides then the base...
thanks
You can't! The sides have area 2 lh+ 2lh and the base has area hw. If the base costs 5 times as much as the sides, the total cost (in units equal to cost of each side) is 2lh+ 2lh+ 5lw.
 

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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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