Question

A rectangular metal tank with open top is to hold 256 cubic feet of liquid. What are the dimensions of the tank that require the least material to build?

Short answer

The dimensions of the tank are 8 feet by 8 feet by 4 feet.

Step-by-step solution

Understand the Problem

We are tasked with finding the dimensions of a rectangular tank with an open top that will require the least amount of material to build, given a volume of 256 cubic feet. The tank dimensions we need to find are length, width, and height. The formula for the volume of a rectangular tank is \( V = lwh \).

Set up the Equations

We know the volume \( V = 256 \) cubic feet, which gives us the equation: \[ lwh = 256. \] We also need to minimize the material used. The surface area \( A \) of the open-top tank is given by: \[ A = lw + 2lh + 2wh. \]

Express One Variable in Terms of Others

From the volume equation \( lwh = 256 \), solve for one variable. For example, solve for \( h \): \[ h = \frac{256}{lw}. \] Substitute this expression into the surface area formula: \[ A = lw + 2l\left(\frac{256}{lw}\right) + 2w\left(\frac{256}{lw}\right). \]

Simplify the Surface Area Expression

Simplify the surface area expression by substituting \( h \):\[ A = lw + \frac{512}{w} + \frac{512}{l}. \] Now, minimize \( A \) with respect to \( l \) and \( w \).

Use Calculus to Minimize the Surface Area

Take the partial derivatives of \( A \) with respect to \( l \) and \( w \), set them equal to zero, and solve the resulting system of equations for \( l \) and \( w \):\[ \frac{\partial A}{\partial l} = w - \frac{512}{l^2} = 0, \]\[ \frac{\partial A}{\partial w} = l - \frac{512}{w^2} = 0. \]

Solve the Derivative Equations

Solve these equations to find \( l \) and \( w \):From \( w = \frac{512}{l^2} \), substitute into the second equation: \[ l^3 = 512, \]solve to find \( l = 8 \). Substituting \( l = 8 \) into \( w = \frac{512}{l^2} \), we find \( w = 8 \).

Determine the Height

Substitute \( l = 8 \) and \( w = 8 \) back into the volume equation to solve for \( h \):\[ h = \frac{256}{8 \times 8} = 4. \]

Verify the Solution

Double-check the calculations by verifying that the volume equation holds: \[ lwh = 8 \cdot 8 \cdot 4 = 256. \] Confirm that the dimensions satisfy the conditions and minimize the surface area.

Key concepts

Rectangular Tank

In this problem, we need to understand what a rectangular tank with an open top is all about. A rectangular tank, as the name suggests, has the shape of a rectangle. Imagine something like a giant box that holds liquid. However, in this case, our tank does not have a lid, so the top is open.

When dealing with tanks like this, you often need to think about things like:

• Length (the longest side)
• Width (the short side, perpendicular to the length)
• Height (how tall the tank is)

These three dimensions define how much the tank can hold inside, which is its volume. In our problem, that volume is given as 256 cubic feet.

Calculus

We use calculus to help us find the best dimensions for the tank. Calculus is a branch of mathematics that allows us to understand changes and optimize solutions. Suppose you want to make our tank in the most material-efficient way, meaning using the least amount of metal.

We apply calculus by finding the derivatives of our equations. Specifically, using the concept of partial derivatives lets us find how a function changes with each of its variables independently. This is extremely useful when you want to optimize different factors, like minimizing surface area.

In this problem, the use of calculus plays the role of guiding us to the optimal dimensions for the rectangular tank while keeping the volume constant at 256 cubic feet.

Volume and Surface Area

The volume and surface area are two crucial mathematical concepts that we need to work out this problem. Let's break them down:

* **Volume**: When we say the tank's volume is 256 cubic feet, it means that's how much space it can hold inside. The formula for volume is straightforward for a rectangle: \[ V = lwh \]This formula tells us that the volume depends on the tank's length, width, and height.

* **Surface Area**: The surface area, on the other hand, tells us how much metal we'd need to make the tank. We only need to count the areas of the sides and the bottom since the top is open. This gives us the formula: \[ A = lw + 2lh + 2wh \]The goal is to minimize this area because we want the tank made with the least material possible.

Understanding these two concepts allows us to set up equations we can solve using calculus to find the optimal dimensions.

Partial Derivatives

Partial derivatives help us solve optimization problems involving multiple variables, like our rectangle's length, width, and height. When we mention partial derivatives, we talk about finding how a function changes when one of those variables changes, while keeping others constant.

In our optimization, we take the partial derivatives of the surface area equation concerning both length \( l \) and width \( w \), and set them to zero. By doing so, we are finding the points where a small change in \( l \) or \( w \) doesn’t lead to an increase or decrease in surface area. This process helps us identify critical points that could potentially minimize the surface area.

For this problem, solving these derivative equations gives us the optimal dimensions for the tank. By setting the partial derivatives to zero and solving, we find the dimensions that make the most efficient use of material while still holding the required volume.