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(1 point) \( \mathrm{A} \) box has a bottom with one edge 2 times as long as the other. If the box has no top and the volume is fixed at \( V \). what dimensions minimize the surface area? dimensions

User PsychoFish
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Final Answer:

The dimensions that minimize the surface area of the box with a fixed volume ( V ) are
\( a = \sqrt[3]{(V)/(2)} \) and
( b = 2\sqrt[3]{(V)/(2)} \), where ( a ) is the shorter edge and ( b ) is the longer edge.

Explanation:

To find the dimensions that minimize the surface area of the box, we need to express the surface area ( S ) in terms of one variable and then apply calculus to find the critical points. The volume ( V ) is fixed, and for a box with no top, the surface area is given by:


\[ S = a^2 + 4ab \]

where ( a ) is the length of the shorter edge, and ( b ) is the length of the longer edge. Since the box has a fixed volume,
\( V = a^2b \). We can express (b ) in terms of ( a ) as
\( b = (V)/(a^2) \). Substitute this expression for ( b) into the surface area formula:


\[ S(a) = a^2 + 4a \left( (V)/(a^2) \right) \]

Simplify
\( S(a) \) to obtain
\( S(a) = a^2 + (4V)/(a) \) . To minimize ( S ), we take the derivative
\( S'(a) \) and set it equal to zero:


\[ S'(a) = 2a - (4V)/(a^2) = 0 \]

Solving for ( a ), we find
\( a = \sqrt[3]{(V)/(2)} \) . Substitute this value back into the expression for ( b ) to get
\( b = 2\sqrt[3]{(V)/(2)} \). These dimensions minimize the surface area for a given volume.

User Invisiblerhino
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