2.10 Q-1

Question Statement

Find two positive integers whose sum is 30 and their product will be maximum.

Background and Explanation

This problem involves optimization, specifically maximizing a product. We are asked to find two positive integers that sum to 30, and we need to determine which pair gives the largest product. To solve this, we’ll use algebraic methods and differentiation to find the maximum point of the function that represents the product.

Solution

Define the variables:

Let the two integers be $x$ and $30 - x$ , where $x$ is one of the integers and $30 - x$ is the other.
Set up the product function:

The product of the two integers can be written as:

f(x) = x(30 - x)

This equation represents the product of the two numbers.

Simplify the function:

Expanding the product function:

f(x) = x(30 - x) = 30x - x^2

Find the derivative of the function:

To maximize the function, we need to find the critical points by differentiating $f(x)$ . The derivative is:

f'(x) = 30 - 2x

Set the derivative equal to zero:

To find the critical points, set the derivative equal to zero:

30 - 2x = 0

Solve for $x$ :

Solving the equation:

2x = 30 \quad \Rightarrow \quad x = 15

Verify the maximum:

To confirm that this value of $x$ gives a maximum, check the second derivative:

f''(x) = -2

Since $f''(x)$ is negative, this indicates that $x = 15$ corresponds to a maximum.

Find the second integer:

The other integer is $30 - 15 = 15$ .

Thus, the two integers that maximize the product are 15 and 15.

Key Formulas or Methods Used

Product function: $f(x) = x(30 - x) = 30x - x^2$
Derivative: $f'(x) = 30 - 2x$
Second derivative test: $f''(x) = -2$

Summary of Steps

Define the integers as $x$ and $30 - x$ .
Write the product function: $f(x) = 30x - x^2$ .
Differentiate to get $f'(x) = 30 - 2x$ .
Set the derivative equal to zero: $30 - 2x = 0$ , solving gives $x = 15$ .
Check the second derivative to confirm a maximum.
The integers are 15 and 15, giving the maximum product.