Maximise
f(x; y) = x
s.t. y =-1/2x*2 + 2x and y =1/2x*2 - 2x + 4
using the level curves method.
Is this point also a stationary point of
the Lagrangian? Why?
This is a very odd problem. Do you mean \(\displaystyle z = f(x,\ y).\)
If the function is as you have specified, it is a function in one variables, and level curves do not apply.
f(x, y) = x implies that y is irrelevant, and that f(x, y) = g(x) = x.
If we are saying that z is a function of x and y, then we are not given sufficient information to deduce level curves.
In any case, if you have one variable subject to two constraints of
equality, then the variable must be equal to both.
\(\displaystyle y = -\ \dfrac{x^2}{2} + 2x \text { and } y= \dfrac{x^2}{2} - 2x + 4 - y \implies\)
\(\displaystyle -\ \dfrac{x^2}{2} + 2x = \dfrac{x^2}{2} - 2x + 4 = 0 \implies x^2 - 4x + 4 = (x - 2)^2 = 0 \implies \)
\(\displaystyle x = 2 \implies y = -\ \dfrac{2^2}{2} + 2 * 2 = -\ 2 + 4 = 2.\)
Maximum occurs at x = 2, y = 2.
