Maximize a^2 + b^2 given a^3 + b^3 = 16

Key concept: This is a constrained optimization problem with a symmetric condition. That symmetry is a big clue: the maximum usually sits on the boundary where one variable takes most of the mass.

Step by step

We want to maximize

$C=a^2+b^2$

subject to

$a^3+b^3=16, \qquad a,b>0.$

Because the constraint and objective are symmetric, we can use the power-mean idea or check endpoints via convexity.

Let $x=a^3$ and $y=b^3$. Then $x+y=16$, and

$C=x^{2/3}+y^{2/3}.$

Since the function $u^{2/3}$ is concave for $u>0$, the sum $x^{2/3}+y^{2/3}$ is maximized when the variables are as unequal as possible, subject to positivity. So the maximum occurs at the boundary case where one variable tends to 0 and the other tends to 16.

Thus,

$a^3\to 16,\quad b^3\to 0$

or vice versa. Then

$C_{\max}=16^{2/3}=\bigl(2^4\bigr)^{2/3}=2^{8/3}=4\sqrt[3]{4}.$

So the maximum value is

$\boxed{4\sqrt[3]{4}}.$

If you want a more formal argument with Lagrange multipliers, you get the same conclusion: the only interior critical point is $a=b=2$, which gives $C=8$, but the larger value is approached when one variable becomes very small and the other approaches $16^{1/3}$.

Pitfall alert

A common slip is to assume the symmetric point $a=b$ must be the maximum. For concave powers like $x^{2/3}$, symmetry is often the minimum, not the maximum. Another trap is forgetting that the problem only requires $a,b>0$, so the boundary can be approached even if $b=0$ is not allowed exactly.

Try different conditions

If the problem had added a lower bound like $a,b\ge 1$, then the boundary argument would change. You would check the endpoints $a=1$ or $b=1$ and compare with the interior point $a=b=2$. In that restricted version the maximum would come from one variable at the allowed minimum and the other adjusted by the cubic constraint.

Further reading

constrained optimization, Lagrange multipliers, power mean