Maximize xy subject to x+y=10 and x,y 0.

Reduce to one variable. Substituting y=10-x gives the objective g(x)=x(10-x)=10x-x2. Maximize the quadratic. This downward parabola is maximized at x=5, giving y=5 and objective value 25.

Maximizing a Product Under a Sum Constraint

Linear Algebra

Optimization

Easy

Maximize

x y

subject to

x + y = 10

and

x, y \geq 0

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Hint 1

Substitute $y = 10 - x$ to reduce the problem to a single-variable quadratic.

Hint 2

The resulting quadratic $x (10 - x)$ is maximized at its vertex.

Show worked solution

1. Reduce to one variable
Substituting $y = 10 - x$ gives the objective $g (x) = x (10 - x) = 10 x - x^{2}$ .
2. Maximize the quadratic
This downward parabola is maximized at $x = 5$ , giving $y = 5$ and objective value $25$ .
$x = y = 5, x y = 25$

Answer: x=y=5, maximum value 25

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Maximizing a Product Under a Sum Constraint

1. Reduce to one variable

2. Maximize the quadratic

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