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Problem 560

# Which 3 -dimensional rectangular box of a given volume $$\mathrm{V}$$ has the least surface area?

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To minimize the surface area of a 3-dimensional rectangular box with a given volume $$V$$, the dimensions should be: Length (l): $$\sqrt{\dfrac{2V}{\sqrt{2V}}}$$ Width (w): $$\sqrt{2V}$$ Height (h): $$\dfrac{V}{\sqrt{\dfrac{2V}{\sqrt{2V}}} \cdot \sqrt{2V}}$$
See the step by step solution

## Step 1: Write the volume and surface area formulas

The volume V of a rectangular box is given by the formula: $$V = lwh$$ where 'l' is the length, 'w' is the width, and 'h' is the height. The surface area S of the box is given by the formula: $$S = 2lw + 2lh + 2wh$$ We want to minimize the surface area given the volume constraint.

## Step 2: Eliminate one variable using the volume constraint

Since the volume V is given, we can rewrite the height as: $$h = \dfrac{V}{lw}$$ Substitute this expression into the surface area formula: $$S(l, w) = 2lw + 2l \cdot \dfrac{V}{lw} + 2w \cdot \dfrac{V}{lw}$$

## Step 3: Simplify the surface area formula

Simplify the expression to have a cleaner formula: $$S(l, w) = 2lw + \dfrac{2V}{w} + \dfrac{2V}{l}$$

## Step 4: Find the critical points

To find the critical points, we need to find the first partial derivatives of the surface area formula with respect to the length and width and set them equal to zero: $$\dfrac{\partial S}{\partial l} = 2w - \dfrac{2V}{l^2} = 0$$ $$\dfrac{\partial S}{\partial w} = 2l - \dfrac{2V}{w^2} = 0$$ Since we have two equations and two unknown variables l and w, we can solve them simultaneously.

## Step 5: Solve for one of the unknown variables

From the first partial derivative equation, we have: $$l^2=\dfrac{2V}{w}$$ From the second partial derivative equation, we have: $$w^2=\dfrac{2V}{l}$$ Now, we can substitute the value of $$l^2$$ from the first equation into the second equation: $$w^2=\dfrac{2V}{ \left(\dfrac{2V}{w} \right)}$$

## Step 6: Solve for width

By solving the equation above, we can find the value of w: $$w^3 = 2V$$ $${w} = \sqrt{2V}$$

## Step 7: Solve for length

Now, substitute the value of w back into the first equation to find l: $$l^2=\dfrac{2V}{ \sqrt{2V}}$$ $$l=\sqrt{\dfrac{2V}{\sqrt{2V}}}$$

## Step 8: Calculate the height

Now that we have the values of l and w, we can use the volume constraint formula to find h: $$h = \dfrac{V}{lw}$$ $$h=\dfrac{V}{\sqrt{\dfrac{2V}{\sqrt{2V}}} \cdot \sqrt{2V}}$$ Now we have found the dimensions of the rectangular box that minimize its surface area for a given volume: Length (l): $$\sqrt{\dfrac{2V}{\sqrt{2V}}}$$ Width (w): $$\sqrt{2V}$$ Height (h): $$\dfrac{V}{\sqrt{\dfrac{2V}{\sqrt{2V}}} \cdot \sqrt{2V}}$$

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