2.10 Q-9

Question Statement

An open tank with a square base and vertical sides is to be constructed to contain a given quantity of water. Find the depth of the tank in terms of the base side length $x$, such that the expense of lining the inside of the tank with lead is minimized.

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Background and Explanation

In this problem, we are asked to minimize the cost of lining the inside of a tank. The cost depends on the surface area to be covered, which in turn depends on the dimensions of the tank. We need to find the depth of the tank in terms of the base side length $x$ that will minimize the surface area. The surface area includes the base and four vertical sides, and we will use calculus to find the optimal dimensions.

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Solution

1. Define the variables:
   • Let $x$ be the side length of the square base.
   • Let $h$ be the depth of the tank.
   • The given quantity of water is the volume of the tank, so the volume $q$ is:

$$q = x^2 \cdot h$$

Solving for $h$ in terms of $x$ and $q$:

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

2. Surface area of the tank: The surface area $S$ is the sum of the area of the base and the areas of the four vertical sides. The area of the base is $x^2$, and the area of the four sides is $4 \cdot x \cdot h$. Thus, the total surface area is:

$$S = x^2 + 4 \cdot x \cdot h$$

Substituting $h = \frac{q}{x^2}$ into the surface area formula:

$$S = x^2 + 4 \cdot x \cdot \frac{q}{x^2}$$

Simplifying:

$$S = x^2 + \frac{q}{x}$$

3. Differentiate the surface area function: To minimize the surface area, we take the derivative of $S$ with respect to $x$:

$$\frac{dS}{dx} = 2x - \frac{4q}{x^2}$$

Setting the derivative equal to zero to find the critical points:

$$2x - \frac{4q}{x^2} = 0$$

Solving for $x$:

$$2x^3 - 4q = 0$$ $$2x^3 = 4q$$ $$x^3 = 2q$$ $$x = (2q)^{\frac{1}{3}}$$

4. Check the second derivative: To confirm that this critical point corresponds to a minimum, we compute the second derivative of $S$:

$$\frac{d^2S}{dx^2} = 2 + \frac{8q}{x^4}$$

Substituting $x = (2q)^{\frac{1}{3}}$:

$$\frac{d^2S}{dx^2} = 2 + 6 = 8$$

Since the second derivative is positive, $x = (2q)^{\frac{1}{3}}$ corresponds to a minimum.

5. Find the depth $h$: Finally, substitute $x = (2q)^{\frac{1}{3}}$ into the equation for $h$:

$$h = \frac{q}{x^2} = \frac{\frac{x^3}{2}}{x^2} = \frac{x}{2}$$

Thus, for the least expense, the depth of the tank is $h = \frac{x}{2}$.

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Key Formulas or Methods Used

• Volume of the tank: $q = x^2 \cdot h$
• Surface area of the tank: $S = x^2 + \frac{q}{x}$
• First derivative test: To find the critical points and minimize the surface area.
• Second derivative test: To confirm that the critical point corresponds to a minimum.

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Summary of Steps

1. Express the depth $h$ in terms of $x$ and $q$: $h = \frac{q}{x^2}$.
2. Write the surface area function: $S = x^2 + \frac{q}{x}$.
3. Differentiate the surface area function: $\frac{dS}{dx} = 2x - \frac{4q}{x^2}$.
4. Set $\frac{dS}{dx} = 0$ and solve for $x = (2q)^{\frac{1}{3}}$.
5. Confirm the minimum using the second derivative: $\frac{d^2S}{dx^2} = 8$.
6. Find the depth $h = \frac{x}{2}$.

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