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?

Answer 1
Without a specific value for the volume, #v#, only a generalized solution can be developed.

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

Let #w = # width of the base; #d =# depth of the base; and #h = # height of the aquarium #C = # cost
#C = # cost of base #+# cost of walls #= (5)( w xx d) +(1)( 2(w+d) xx h)#
Note that the minimum will be achieved when the width and the depth are equal. (I think this should be obvious but re-post if it isn't). So we are only interested in the case (substituting #w# for #d#)
#C = 5w^2 + 4wh#
The volume in this case can be expressed as #v = w^2h#
So if we re-write our Cost equation as #C = 5w^2 + (4w^2h)/w#
This becomes #C = 5w^2 + 4vw^(-1)#

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

#(d C)/(dw) = 10w - 4vw^(-2)#
Once a value has been established for the volume (#v#) this equation can be solved for #w# and via substitution for #d# and #h#.
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Answer 2

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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Answer from HIX Tutor

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.

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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