A box of volume 8 m^3 with a square top and bottom is constructed out of two types of metal. The metal for the top and bottom costs $70/m^2 and the metal for the sides costs $40/m^... A box of volume 8 m^3 with a square top and bottom is constructed out of two types of metal. The metal for the top and bottom costs $70/m^2 and the metal for the sides costs $40/m^2. Find the dimensions of the box that minimize the total cost: Length of the sides of the square base and height of the box. Read more

Steps to Solve

1. Set up the volume constraint

Let the side length of the square base be $x$ and the height of the box be $h$. The volume of the box is given by the formula:

$$ V = x^2 h $$

We know the volume is a constant value, say $V_0$. Thus, we can express $h$ in terms of $x$:

$$ h = \frac{V_0}{x^2} $$

2. Define the cost function

Next, we need to determine the cost of materials used to construct the box. Let's assume the cost per square unit of the base is $C_b$ and the cost per square unit of the sides is $C_s$. The surface area of the box comprises:

• Area of the base: $x^2$
• Area of the four sides: $4(xh)$

Thus, the total cost function $C$ becomes:

$$ C = C_b x^2 + C_s(4xh) $$

3. Substitute for height in the cost function

Now, substitute $h$ into the cost function to express everything in terms of $x$:

$$ C = C_b x^2 + C_s \left(4x \cdot \frac{V_0}{x^2}\right) $$

This simplifies to:

$$ C = C_b x^2 + \frac{4C_s V_0}{x} $$

4. Differentiate the cost function

To minimize the cost, we need to take the derivative of the cost function with respect to $x$:

$$ \frac{dC}{dx} = 2C_b x - \frac{4C_s V_0}{x^2} $$

5. Find critical points

Set the derivative equal to zero to find critical points:

$$ 2C_b x - \frac{4C_s V_0}{x^2} = 0 $$

Rearranging gives:

$$ 2C_b x^3 = 4C_s V_0 $$

Now solve for $x$:

$$ x^3 = \frac{2C_s V_0}{C_b} $$

Thus:

$$ x = \left( \frac{2C_s V_0}{C_b} \right)^{1/3} $$

6. Calculate height using found $x$

Using the value of $x$, we can find the height $h$ using our earlier volume constraint:

$$ h = \frac{V_0}{x^2} = \frac{V_0}{\left( \frac{2C_s V_0}{C_b} \right)^{2/3}} $$

This simplifies to:

$$ h = \frac{C_b^{2/3} V_0^{1/3}}{(2C_s)^{2/3}} $$