If a two-variable function under some constraint has only one critical point, can I classify the point as an extrema (global)?

Question

Find the critical points of the function $z=x^2+y^2$ under the constraint $\frac{x}{4}+\frac{y}{3}=1$.

Hey everyone.

I used the Lagrange multiplier to find the critical point $(\frac{36}{25}, \frac{48}{25})$. Now, I am not sure whether to use the second derivative test to classify the point, since I don't know if the test holds under this constraint.

I would appreciate your help, thank you :)

Answer 1

The constraint defines a line $\ell$ in the $(x,y)$-plane. Geometric qualitative considerations tell us that the function $f(x,y):=x^2+y^2$ will take a global minimum on $\ell$, and this global minimum is then a local minimum as well, hence will be brought to the fore by Lagrange's method. Lagrange's method found one point; hence this is the point you are looking for.

Answer 2

$3x=12-4y$

$$9z=(3x)^2+9y^2=(12-4y)^2+9y^2=144+25y^2-96y=144+25\left(y-\dfrac{48}{25}\right)^2-\dfrac{48^2}{25}$$

$$\ge144-\dfrac{48^2}{25}$$