Problem 1 - Calculus for Biology and Medicine

Question

$$ \begin{array}{l} \text { Find the smallest perimeter possible for a rectangle whose area }\\\ \text { is } 25 \mathrm{in} .^{2} \end{array} $$

Step-by-Step Solution

Answer

The smallest perimeter is 20 inches, achieved when the rectangle is a square.

Step 1: Define the Relationship between Dimensions and Area

For a rectangle with length \( l \) and width \( w \), the area \( A \) is given by \( A = l \times w \). We know the area is 25 square inches, so \( l \times w = 25 \).

Step 2: Define the Perimeter Formula

The perimeter \( P \) of a rectangle is given by the formula \( P = 2l + 2w \). We want to minimize this perimeter given that \( l \times w = 25 \).

Step 3: Express One Variable in Terms of the Other

Using the area relationship, express one variable in terms of the other: \( w = \frac{25}{l} \). Substitute this expression into the perimeter formula.

Step 4: Substitute and Simplify the Perimeter Function

Substitute \( w = \frac{25}{l} \) into the perimeter equation: \[ P = 2l + 2\left(\frac{25}{l}\right) = 2l + \frac{50}{l}. \]

Step 5: Find the Derivative of the Perimeter Function

Differentiate \( P = 2l + \frac{50}{l} \) with respect to \( l \). The derivative \( P' \) is given by: \[ P' = 2 - \frac{50}{l^2}. \]

Step 6: Find the Critical Points

Set \( P' = 0 \) to find the critical points: \[ 2 - \frac{50}{l^2} = 0. \] Solve for \( l \): \[ 2 = \frac{50}{l^2} \Rightarrow l^2 = 25 \Rightarrow l = 5. \]

Step 7: Verify and Calculate the Dimensions

Since \( l = 5 \), substitute back to find \( w = \frac{25}{5} = 5 \). The rectangle is a square with sides of 5 inches.

Step 8: Calculate the Minimum Perimeter

Substitute \( l = 5 \) and \( w = 5 \) into the perimeter formula: \[ P = 2(5) + 2(5) = 20. \] The minimum perimeter is 20 inches.

Key Concepts

Perimeter of a Rectangle

The perimeter of a rectangle is the total distance around its outer edge. This measurement is calculated by adding together the lengths of all four sides.
The formula to find the perimeter, \( P \), of a rectangle simply uses its length, \( l \), and width, \( w \):

• The perimeter formula is \( P = 2l + 2w \).
• This equation can be understood as the sum of twice the length plus twice the width.

Understanding how to calculate perimeter is essential in optimization problems because it can help determine how measurements like length and width should relate to achieve a desired outcome. In problems where minimizing or maximizing the perimeter is essential, knowing this formula helps set the scene.

Derivative

A derivative measures how a function changes as its input changes. In mathematical terms, it's the slope of the function at any given point.
For optimization problems, derivatives are powerful tools because they help identify points where peaks (maxima) or valleys (minima) occur.
When working on finding the minimum perimeter of a rectangle, we use the derivative to understand how changes in one dimension affect the perimeter.

• For the function \( P = 2l + \frac{50}{l} \), its derivative \( P' \) helps us pinpoint critical points where the function might have local minima or maxima.
• By finding where the derivative equals to zero, \( P' = 0 \), we can locate these critical points.

In essence, the derivative becomes a crucial tool in figuring out where the minimal perimeter occurs relative to the rectangle's dimensions.

Critical Points

Critical points are important in optimization as they indicate where a function can have a local maximum or minimum. In the context of minimizing the perimeter of a rectangle, finding these points tells us where the perimeter will be smallest.
When using a derivative, the critical points are found by setting the derivative equal to zero and solving for the variable.

• In this problem, we found the critical point by setting \( P' = 2 - \frac{50}{l^2} = 0 \).
• Solving this gives \( l^2 = 25 \), leading to \( l = 5 \) as a critical point.

Identifying these critical points is crucial because they tell us precisely how the rectangle’s dimensions should be adjusted to achieve the minimal perimeter while satisfying the area constraint. It shows when both length and width should ideally be equal, leading to a square.

Area Constraint

An area constraint limits the possible dimensions of a shape to maintain a specific area. In our problem, the area of the rectangle is fixed at 25 square inches. This constraint is essential because it directly informs how changes in one dimension of the rectangle affect the other dimensions.
When tackling optimization problems, maintaining this constraint ensures our solutions are valid and practical.

• The area formula \( A = l \times w = 25 \) is vital for setting up this constraint.
• Using this equation, if either \( l \) or \( w \) is known, the other can be calculated as \( w = \frac{25}{l} \).

By considering the area constraint in conjunction with the perimeter formula, we can craft a function that reflects both desired perimeter minimization and the constant area, leading to an optimized solution.