A rectangular box has a volume of 16 cm^3 and has a square base. Find the dimensions of the box which minimize the surface area of the box.

Answer 1

This is a classic optimization problem. Let's break it down step by step.

Step 1: Define Variables

Let 'x' be the length of a side of the base of the box. Let 'h' represent the height of the box. Since it has a square base, the box has dimensions x × x × h.

Step 2: Volume Constraint

Optimization problems in Calculus often have a constraint. In this problem, we are told that the volume of the box must be 16 cm³. Let's write this formally with an equation, where 'V' is the volume of the box, and solve for 'h'. The volume of a rectangular prism is the product of its length, width, and height.

V = x * x * h

V = x²h

16 = x²h

h = 16 / x²

Step 3: Surface Area Function

Let's write a function that represents the surface area of the box. To calculate the surface area, we need the area of all sides of the box. Since it has a square base, the top and bottom faces will be x × x in area. The four remaining sides of the box will be x × h in area. Here is the surface area 'S' as a function of x and h.

S = 2(x * x) + 4(x * h)

S = 2x² + 4xh

Let's substitute the result from step 2 into this equation so that our surface area is only a function of 'x'.

S(x) = 2x² + 4x(16 / x²)

S(x) = 2x² + 64 / x

Step 4: Optimization

Now that we have a function for the surface area of the constrained box, we can optimize it. To do this, we will take the derivative of S(x) and set it equal to zero. This works because S'(x) = 0 represents the location of potential minimums and maximums, also known as local extrema. Let's use the power rule and set it equal to zero.

S'(x) = 4x - 64 / x²

S'(x) = 0

4x - 64 / x² = 0

4x = 64 / x²

4x³ = 64

x³ = 16

x = ∛(16)

Step 5: Confirmation of Relative Minimum/Maximum

We can use one of two tests to determine if x = ∛(16) is a relative minimum or a relative maximum. Only one is necessary. Let's do the First Derivative Test.

Let's create a sign chart for S'(x). For values of x less than x = ∛(16), such as x = 1, S'(x) is negative. For values of x greater than x = ∛(16), such as x = 4, S'(x) is positive.

S'(1) = 4(1) - 64 / (1)² = 4 - 64 = -60

S'(4) = 4(4) - 64 / (4)² = 16 - 4 = 12

Therefore, the sign of the derivative switches from negative to positive over the critical point, x = ∛(16). This indicates that the critical point at x = ∛(16) is the location of a relative minimum.

We could also use the Second Derivative Test. Taking the second derivative of the surface area function, we get this.

S''(x) = 4 + 128 / x³

Remember that 'x' represents the base side length. That means that x must be greater than zero. Analyzing S''(x), we can see that it will also only be greater than zero. This means that S(x) is concave up. This also indicates that the critical point at x = ∛(16) is the location of a relative minimum.

Step 6: Answering the Problem

Now that we know that the surface area is minimized when x = ∛(16), we need to calculate all of its dimensions. Let's start with 'h'. From step 2, we found a relationship between 'h' and 'x'.

h = 16 / x²

h = 16 / (∛(16))²

h = ∛(16)

The optimal dimensions of a box with volume 16 cm³ that minimize surface area are approximately 2.52 cm × 2.52 cm × 2.52 cm.