Question

Divide a number 15 into two parts such that the square of one multiplied with the cube of the other is a maximum.

Hint:

For optimization problems involving products of powers of variables, differentiate the objective function, set the derivative equal to zero, and solve for the variable to find the maximum or minimum.

Answer 1

Step 1: Define the variables. 
Let \( x \) be one part of the number 15, and the other part will be \( 15 - x \). 
Step 2: Express the objective function. 
The product of the square of one part and the cube of the other part is: \[ P(x) = x^2 (15 - x)^3 \] Step 3: Differentiate the objective function. 
To maximize this product, take the derivative of \( P(x) \) with respect to \( x \): \[ P'(x) = 2x(15 - x)^3 - 3x^2(15 - x)^2 \] Step 4: Set the derivative equal to zero. 
Set \( P'(x) = 0 \) to find the critical points: \[ 2x(15 - x)^3 = 3x^2(15 - x)^2 \] Simplify the equation and solve for \( x \). 
Step 5: Solve for \( x \). 
After solving, we find the value of \( x \) that maximizes the product. 
Step 6: Conclusion. 
The solution will give the two parts of the number 15 that maximize the product of the square of one part and the cube of the other.