Find the maximum value of f(x,y) = x2 + y2 subject to the constraint x+y = 1.

We are tasked with finding the maximum value of \( f(x, y) = x^2 + y^2 \), subject to the constraint \( x + y = 1 \).

### Step 1: Express one variable in terms of the other using the constraint.
From the constraint \( x + y = 1 \), we can solve for \( y \) in terms of \( x \):
\[
y = 1 - x
\]

### Step 2: Substitute into the objective function.
Substitute \( y = 1 - x \) into \( f(x, y) = x^2 + y^2 \):
\[
f(x, 1 - x) = x^2 + (1 - x)^2
\]
Now simplify the expression:
\[
f(x, 1 - x) = x^2 + (1 - 2x + x^2) = x^2 + 1 - 2x + x^2 = 2x^2 - 2x + 1
\]

### Step 3: Find the critical points.
To find the maximum, we take the derivative of \( f(x, 1 - x) = 2x^2 - 2x + 1 \) with respect to \( x \) and set it equal to zero:
\[
\frac{d}{dx}(2x^2 - 2x + 1) = 4x - 2
\]
Set the derivative equal to zero to find the critical points:
\[
4x - 2 = 0
\]
\[
x = \frac{1}{2}
\]

### Step 4: Calculate the corresponding value of \( y \).
Using the constraint \( x + y = 1 \), if \( x = \frac{1}{2} \), then:
\[
y = 1 - \frac{1}{2} = \frac{1}{2}
\]

### Step 5: Calculate the maximum value of \( f(x, y) \).
Substitute \( x = \frac{1}{2} \) and \( y = \frac{1}{2} \) into the original function:
\[
f\left( \frac{1}{2}, \frac{1}{2} \right) = \left( \frac{1}{2} \right)^2 + \left( \frac{1}{2} \right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}
\]

### Conclusion:
The maximum value of \( f(x, y) = x^2 + y^2 \) subject to the constraint \( x + y = 1 \) is \( \frac{1}{2} \).

Do you need further details or have any questions?

Here are 5 related questions to expand your understanding:
1. How do you find the minimum value of \( f(x, y) = x^2 + y^2 \) under the same constraint?
2. What happens to the maximum if the constraint is changed to \( x + y = 2 \)?
3. Can you solve this problem using the method of Lagrange multipliers?
4. How does the function \( f(x, y) = x^2 + y^2 \) behave for different types of constraints?
5. How would the solution change if \( f(x, y) \) was \( f(x, y) = x^2 - y^2 \) with the same constraint?

**Tip**: To optimize functions under constraints, it's useful to express one variable in terms of the other, simplifying the problem to one variable.

Find the maximum value of f(x,y) = x^2 + y^2 subject to the constraint x + y = 1.

Discover how to find the maximum value of the function f(x, y) = x^2 + y^2 with the constraint x + y = 1. This detailed step-by-step guide covers algebraic substitution, derivatives, and finding critical points. Ideal for students in Grades 10-12.

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