# How do you find the dimensions of the aquarium that minimize the cost of the materials if the base of an aquarium with volume v is made of slate and the sides are made of glass and the slate costs five times as much (per unit area) as glass?

We need to develop an expression for the cost of materials based on one of the dimensions.

To find the minimum cost, we take the derivative of this and set the result to zero.

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To minimize the cost, you can use calculus to find the dimensions. Let ( x ) be the length of the base and ( y ) be the height of the aquarium. The volume ( V ) is given by ( V = x \cdot x \cdot y = x^2y ). The cost ( C ) is given by ( C = 5x^2 + 2xy ).

To minimize ( C ), differentiate ( C ) with respect to ( x ), set it equal to zero, and solve for ( x ).

[ \frac{dC}{dx} = 10x + 2y = 0 ]

Solve for ( y ) in terms of ( x ) from the volume equation:

[ y = \frac{V}{x^2} ]

Substitute ( y ) in terms of ( x ) into the equation from differentiating ( C ), and solve for ( x ). Then use this value of ( x ) to find ( y ) and minimize the cost ( C ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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