The sum of two number is 20 what are the numbers if the product of the square of one and the cube of the other is maximum ?

Options

A. 6,14
B. 15,5
C. 12,8
D. 10,10

Explanation

To solve the problem step-by-step, we want to maximize the product of the square of one number and the cube of the other, given that the sum of the two numbers is 20.

### Step 1: Define the Variables
Let the two numbers be \( x \) and \( y \). According to the problem, we have:
\[ x + y = 20 \]

### Step 2: Express One Variable in Terms of the Other
From the equation \( x + y = 20 \), we can express \( y \) in terms of \( x \):
\[ y = 20 - x \]

### Step 3: Define the Product Function
We need to maximize the product \( P \) defined as:
\[ P = x^2 \cdot y^3 \]
Substituting \( y \) from Step 2, we get:
\[ P = x^2 \cdot (20 - x)^3 \]

### Step 4: Differentiate the Product Function
To find the maximum value of \( P \), we differentiate it with respect to \( x \):
\[ P = x^2 \cdot (20 - x)^3 \]
Using the product rule:
\[ \frac{dP}{dx} = 2x(20 - x)^3 + x^2 \cdot 3(20 - x)^2(-1) \]
This simplifies to:
\[ \frac{dP}{dx} = 2x(20 - x)^3 - 3x^2(20 - x)^2 \]

### Step 5: Set the Derivative to Zero
To find the critical points, set the derivative equal to zero:
\[ 2x(20 - x)^3 - 3x^2(20 - x)^2 = 0 \]
Factoring out common terms:
\[ x(20 - x)^2(2(20 - x) - 3x) = 0 \]
This gives us:
1. \( x = 0 \)
2. \( 20 - x = 0 \) (i.e., \( x = 20 \))
3. \( 2(20 - x) - 3x = 0 \)

Solving the third equation:
\[ 40 - 2x - 3x = 0 \implies 40 = 5x \implies x = 8 \]

### Step 6: Find Corresponding Values of \( y \)
Using \( x = 8 \):
\[ y = 20 - x = 20 - 8 = 12 \]

### Step 7: Verify Maximum Using Second Derivative Test
To confirm that this is a maximum, we can check the second derivative:
\[ \frac{d^2P}{dx^2} \]
We can evaluate this at \( x = 8 \) to check if it is negative (indicating a maximum).

### Conclusion
The two numbers that maximize the product of the square of one and the cube of the other, given that their sum is 20, are:
\[ \boxed{8 \text{ and } 12} \]