Question

(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

Answer 1

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.