# $f$ does not have extrema at the Lagrange multiplier value

I have the function $2x^2+y$ and one constraint $x-y^2=1$ and want to find maximum value by lagrange multiplier. Intuitively, I see the point $(2,1)$ satisfies $c$ and have value of $f(2,1)=9$.

Using Lagrange function $f(z)-\lambda c(z)$, I get $\lambda^3-4\lambda^2-1$ thus $\lambda=4.06$. However, when plug this into the derivatives of $x$ and $y$ I got different values in particular $x=\frac{4.06}{4}=1.015$ which is much lower than $2$. Am I missing something here?

• Actually, $\lambda\approx-4.06$. – Michael Hoppe Nov 18 '13 at 20:55
• @MichaelHoppe Thanks. plugging $\lambda^3-4\lambda^2-1$ into the calc here 1728.org/cubic.htm shows me 4.06 – seteropere Nov 19 '13 at 0:06

Note that your function is unbounded. Your constraint gives $x = 1+y^2$, and plugging this into the objective, you get $2\,(1+y^2)^2 + y$. Now, it is easy to see, that this can attain arbitrary large values. This means that your problem has no global maximum, but it may have a local one.

Further, since the gradient of your constraint is $(1, -2\,y)$, the LICQ is satisfied at every feasible point. This means that every local extremum (may it be a maximum or a minimum) can be found by the Lagrange multiplier method.

As you already found, there is only one solution to Lagrange's method. Moreover, it is easy to see that this is a local minimum (consider, e.g., the second derivative of the reduced objective $2\,(1+y^2)^2+y$ which is positive for $y = -0.123$.

Putting everything together, we find that your problem has no local maximum (and no global one).

Nobody cares about the value of $\lambda$. Eliminating $\lambda$ yields $1+8xy=0$ and $x-y^2=1$, so solve $8y^3+8y+1=0$. Good luck.

• Sorry I didn't get what you mean by eliminating $\lambda$. I have $x=\frac{\lambda}{4}$ and $y=\frac{-1}{2\lambda}$. I need to find the value of $\lambda$ so I can substitute it in $x$ and $y$. I tried subtituting $\lambda=4x$ and $\lambda=\frac{-1}{2y}$ in $\lambda^3-4\lambda^2-1=0$. – seteropere Nov 18 '13 at 19:47
• Who cares about $\lambda$? We want to find $x$ and $y$. So our goal is to get rid of $\lambda$, in our case $\lambda=-4x$ (note: $x=-\lambda/4$, not $x=\lambda/4$ as you stated.). Just substitute $\lambda=-4x$ in $1-2\lambda y=0$ to achieve $1+8xy=0$. From the constrain we know $x=1+y^2$, hence solve $8y^3+8y+1=0$. – Michael Hoppe Nov 18 '13 at 20:05
• Thanks. I am multiplying the constraint by $-\lambda$ thats why $x=\lambda/4$. One final question, the resulted points are $(1.015,-.123)$ with $f(1.015,-.123)=1.9$. Clearly, these are not maximum points, is this mean they are minimum? – seteropere Nov 19 '13 at 0:27