Maximize xy Subject to x² + xy + y² = 1

Interpreting the constraint as $x^2+xy+y^2=1$, you can find the maximum of $f(x,y)=xy$ with a one-line inequality. The key is to use a square that is always nonnegative, then check when equality occurs.

A direct proof of the maximum

For every pair of real numbers $x$ and $y$,

$$(x-y)^2\ge 0.$$

Expanding and rearranging gives

$$x^2+y^2\ge 2xy.$$

Now use the constraint to replace $x^2+y^2$ with $1-xy$:

$$1-xy\ge 2xy.$$

Therefore,

$$xy\le \frac{1}{3}.$$

This proves that no feasible pair can give a value larger than $1/3$. To show that this bound is the actual maximum, find a feasible pair that reaches it. Equality in the square inequality holds when $x=y$. Substituting $y=x$ into the constraint gives $3x^2=1$, so

$$x=y=\frac{1}{\sqrt{3}}\quad\text{or}\quad x=y=-\frac{1}{\sqrt{3}}.$$

At either point, $xy=1/3$. Thus the maximum is $1/3$, attained at both points.

Check the equality cases

For the positive choice, $x=y=1/\sqrt{3}$, the constraint evaluates to

$$\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1,$$

and the objective is $xy=1/3$. For the negative choice, $x=y=-1/\sqrt{3}$, the squared terms are still $1/3$ each and the product is still $1/3$, so this point also satisfies the constraint and reaches the bound.

The two points matter: the objective depends on the product, so changing both signs leaves $xy$ unchanged. The constraint also remains satisfied because its three terms are all unchanged when both variables change sign.

Lagrange multiplier check

The inequality proves the global upper bound directly. You can also check the possible extrema with Lagrange multipliers. Let $f(x,y)=xy$ and $g(x,y)=x^2+xy+y^2$. At a constrained stationary point,

$$\nabla f=\lambda\nabla g,$$

so

$$y=\lambda(2x+y),\qquad x=\lambda(x+2y).$$

Subtracting the second equation from the first gives $(1+\lambda)(y-x)=0$. There are two cases. If $y=x$, the constraint gives $x=y=\pm1/\sqrt{3}$ and $xy=1/3$. If $\lambda=-1$, the first equation implies $y=-x$; the constraint then gives $x=\pm1$, $y=\mp1$, and $xy=-1$.

The constraint describes a closed, bounded ellipse, so the objective has a largest and a smallest value on it. The multiplier check finds the maximum $1/3$ and the minimum $-1$. It also shows why the points with opposite signs, although they give a product with larger absolute value, are not the maximum.

Common mistakes

• Choosing the product with the largest absolute value. The question asks for the largest value, not the largest magnitude. The value $-1$ is smaller than $1/3$.
• Dropping the $xy$ term in the constraint. The equation is not $x^2+y^2=1$; it is $x^2+xy+y^2=1$.
• Checking only one equality point. Both $(1/\sqrt{3},1/\sqrt{3})$ and $(-1/\sqrt{3},-1/\sqrt{3})$ attain the maximum.
• Stopping after finding an upper bound. A bound alone does not prove it is the maximum. You must also show that a feasible pair attains it.