Question

find the maximum value of z=6x-3x^2+2y

(subject to y-x^2=2

if condition changes to y-x^2=3, how much will z increase?

Answer 1

The maximization problem is:

max 6x – 3x +2y s.t. y - x2 = 2

Setting up the Lagrange:
$\small \mathcal{L}=6x-3x^2+2y+\lambda(y-x^2-2)$
The FOCs are:
$\small \frac{\partial \mathcal{L}}{\partial x}=6-6x-2\lambda x=0$
$\small \frac{\partial \mathcal{L}}{\partial y}=2+\lambda=0$
$\small y-x^2=2$
Dividing the first two FOCs:
$\small \frac{6-6x}{2}=\frac{2\lambda x}{-\lambda}\rightarrow x=3$
Substituting this into the third FOC:
$\small y-3^2=2\rightarrow y=11$
The solution to the maximization problem is:
$\small (x,y)=(3,11)$

If the condition changes to $\small y-x^2=3$ ,
All the FOCs will remain the same except the last one. The last FOC will change to:
$\small y-x^2=3$
Since the first two FOCs do not change, the solution to the x coordinate will not change. Substituting the value of x into the new FOC:
$\small y-3^2=3\rightarrow y=12$
The solution to the new maximization problem is:
$\small (x,y)=(3,12)$