Optimal Dimensions Rectangular Tank Minimum Surface Area

Introduction

Hey guys! Let's dive into a super cool problem today – a classic optimization challenge. We're going to figure out how to build a rectangular tank with a square base, an open top, and a whopping volume of 19,652 cubic feet, all while using the least amount of sheet steel possible. In other words, we want to minimize the surface area. This is a fantastic real-world application of calculus, showing how math can help us be efficient and save resources. So, buckle up, and let’s get started!

Problem Statement

Let's restate the problem clearly. We need to construct a rectangular tank that meets specific criteria:

• It has a square base.
• The top is open (no lid).
• The volume must be exactly 19,652 ft³. Our main goal is to find the dimensions (the side length of the base and the height) that will give us the minimum surface area. Think about it – using less material means lower costs and less waste, which is always a win!

Setting up the Equations

Okay, let's get mathematical! To solve this, we need to translate the problem into equations. First, let’s define our variables:

• Let x be the length of a side of the square base (in feet).
• Let h be the height of the tank (in feet).

Now, let’s express the volume and surface area in terms of x and h.

Volume

The volume V of a rectangular tank with a square base is given by:

V = x²h

We know that the volume is 19,652 ft³, so we have:

x²h = 19,652

This equation is our constraint. It tells us that the dimensions x and h must satisfy this relationship to give us the required volume.

Surface Area

Now, let's think about the surface area A. Since the tank has an open top, we only need to consider the base and the four sides. The area of the square base is x², and the area of each of the four sides is xh. Therefore, the total surface area is:

A = x² + 4xh

This is the function we want to minimize. Our goal is to find the values of x and h that make A as small as possible, while still satisfying the volume constraint.

Minimizing the Surface Area

Here's where the magic of calculus comes in! We have two variables, x and h, but we want to minimize a function of one variable. To do this, we'll use our volume constraint to eliminate one of the variables.

Expressing h in terms of x

From the volume equation x²h = 19,652, we can solve for h:

h = 19,652 / x²

Substituting into the Surface Area Equation

Now, we substitute this expression for h into the surface area equation:

A = x² + 4x(19,652 / x²) = x² + 4(19,652) / x = x² + 78,608 / x

We now have the surface area A as a function of a single variable, x: A(x) = x² + 78,608 / x. This is fantastic! Now we can use calculus to find the minimum.

Finding the Critical Points

To minimize A(x), we need to find its critical points. These are the points where the derivative is either zero or undefined. Let's find the derivative A'(x):

A'(x) = d/dx (x² + 78,608 / x) = 2x - 78,608 / x²

To find the critical points, we set A'(x) = 0:

2x - 78,608 / x² = 0

Multiplying through by x² to clear the fraction, we get:

2x³ - 78,608 = 0

2x³ = 78,608

x³ = 39,304

Taking the cube root of both sides, we find:

x = ∛39,304 = 34 ft

So, we have one critical point at x = 34. We also need to consider where A'(x) is undefined, which is at x = 0. However, x = 0 doesn’t make sense in our context (we can't have a tank with a base side length of 0), so we can ignore it.

Verifying the Minimum

We need to make sure that x = 34 actually gives us a minimum surface area. We can use the second derivative test for this. Let's find the second derivative A''(x):

A''(x) = d²/dx² (x² + 78,608 / x) = d/dx (2x - 78,608 / x²) = 2 + 2(78,608) / x³

Now, let's evaluate A''(x) at x = 34:

A''(34) = 2 + 2(78,608) / (34)³ = 2 + 157,216 / 39,304 = 2 + 4 = 6

Since A''(34) = 6 > 0, the function A(x) has a local minimum at x = 34. This confirms that we've found the value of x that minimizes the surface area.

Finding the Height

Now that we have the optimal value for x, we can find the corresponding height h using the equation h = 19,652 / x²:

h = 19,652 / (34)² = 19,652 / 1156 = 17 ft

The Dimensions of the Tank

Alright, we've done it! We've found the dimensions of the tank that minimize the surface area. Here they are:

• Base side length: x = 34 ft
• Height: h = 17 ft

So, to build the tank with the least amount of sheet steel, it should have a square base with sides of 34 feet and a height of 17 feet.

The Minimum Surface Area

Just for kicks, let's calculate the minimum surface area using these dimensions:

A = x² + 4xh = (34)² + 4(34)(17) = 1156 + 2312 = 3468 ft²

So, the minimum surface area required to build the tank is 3468 square feet.

Conclusion

This problem demonstrates a classic application of calculus in optimization. We used our knowledge of derivatives and critical points to find the dimensions of a tank that minimize its surface area while maintaining a specific volume. This type of problem is not just a mathematical exercise; it has practical implications in engineering and design, where minimizing material usage can lead to significant cost savings and resource efficiency. Awesome, right?

I hope you found this explanation helpful and insightful! Keep practicing these types of problems, and you'll become a master of optimization in no time. Until next time, happy problem-solving!

The tank with the minimum surface area has a height of 17 ft