Find three numbers whose sum is 24 and whose sum of squares is a minimum.

We are tasked with finding three numbers whose sum is 24, and the sum of their squares is minimized.

### Step 1: Define Variables
Let the three numbers be \( x \), \( y \), and \( z \).

We are given that:
\[
x + y + z = 24
\]
and we need to minimize the sum of their squares:
\[
S = x^2 + y^2 + z^2
\]

### Step 2: Express \( z \) in Terms of \( x \) and \( y \)
From the constraint \( x + y + z = 24 \), we can solve for \( z \):
\[
z = 24 - x - y
\]

### Step 3: Substitute into the Sum of Squares
Substitute \( z = 24 - x - y \) into the equation for \( S \):
\[
S = x^2 + y^2 + (24 - x - y)^2
\]
Expand the square:
\[
S = x^2 + y^2 + (24 - x - y)^2 = x^2 + y^2 + (576 - 48x - 48y + x^2 + 2xy + y^2)
\]
Simplify the expression:
\[
S = 2x^2 + 2y^2 + 2xy - 48x - 48y + 576
\]

### Step 4: Find the Critical Points
To minimize \( S \), we take the partial derivatives of \( S \) with respect to \( x \) and \( y \), and set them equal to zero.

First, compute the partial derivatives:
\[
\frac{\partial S}{\partial x} = 4x + 2y - 48
\]
\[
\frac{\partial S}{\partial y} = 4y + 2x - 48
\]

Set both partial derivatives equal to zero:
\[
4x + 2y - 48 = 0
\]
\[
4y + 2x - 48 = 0
\]

### Step 5: Solve the System of Equations
From both equations, we have:
\[
4x + 2y = 48 \quad \text{(Equation 1)}
\]
\[
4y + 2x = 48 \quad \text{(Equation 2)}
\]

Both equations are identical, so we can solve for \( x \) and \( y \) by rearranging Equation 1:
\[
2x + y = 24
\]
\[
y = 24 - 2x
\]

### Step 6: Substitute into the Constraint
Now, substitute \( y = 24 - 2x \) into the constraint \( x + y + z = 24 \):
\[
x + (24 - 2x) + z = 24
\]
Simplify:
\[
x + 24 - 2x + z = 24
\]
\[
-x + z = 0
\]
\[
z = x
\]

### Step 7: Conclusion
We have found that \( y = 24 - 2x \) and \( z = x \). Therefore, the three numbers are \( x \), \( 24 - 2x \), and \( x \).

Now, to minimize the sum of their squares, we can check the values of \( x \). Since we are minimizing \( S = 2x^2 + 2y^2 + 2xy - 48x - 48y + 576 \), we find that the minimum value occurs when all three numbers are equal. Thus:

\[
x = 8, \quad y = 8, \quad z = 8
\]

So, the three numbers are **8, 8, and 8**. Their sum is 24, and their sum of squares is minimized.

---

Would you like to go over any part of the solution or need further clarification?

### 5 Follow-up Questions:
1. How do you find critical points for functions with multiple variables?
2. What happens when the numbers are not equal, and why is equality optimal here?
3. How do partial derivatives help in optimization problems?
4. Can this method be used for more than three variables?
5. What would happen if we had additional constraints?

**Tip**: When dealing with optimization problems under constraints, it's useful to rewrite the constraint and substitute into the objective function to simplify the process.

Learn how to find three numbers whose sum is 24 and whose sum of squares is minimized through optimization techniques. This problem involves algebra, partial derivatives, and solving systems of equations to find the optimal values. Suitable for high school students in Grades 11-12.

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