Pessimist Singularitarian

#12
January 3rd, 2018,
15:04
As a follow-up, I would begin by adding the two equations to eliminate $x$ and we obtain:

$ \displaystyle 180-2y=0\implies y=90$

Now, substituting for $y$ into the first equation, there results:

$ \displaystyle 100+2(90)-2x=0\implies x=140$

And so our critical point is:

$ \displaystyle (x,y)=(140,90)$

Now, in order to use the second partials test for relative extrema, we need to compute:

$ \displaystyle P_{xx}(x,y)=-2$

$ \displaystyle P_{yy}(x,y)=-4$

$ \displaystyle P_{xy}(x,y)=2$

And we find:

$ \displaystyle D(x,y)=P_{xx}(x,y)P_{yy}(x,y)-\left(P_{xy}(x,y)\right)^2=8-4=4>0$

And so we conclude that the critical point is at the global maximum, which is:

$ \displaystyle P(140,90)=5600$

For the second problem, we have the objective function:

$ \displaystyle U(x,y)=xy$

Subject to the constraint:

$ \displaystyle g(x,y)=5x+10y-100=0$

Using Lagrange multipliers, we obtain the system:

$ \displaystyle y=\lambda(5)$

$ \displaystyle x=\lambda(10)$

We find this implies:

$ \displaystyle x=2y$

Substituting for $x$ into the constraint, we have:

$ \displaystyle 5(2y)+10y-100=0\implies (x,y)=(10,5)$

We find:

$ \displaystyle U(10,5)=50$

Picking another point on the constraint, such as $(12,4)$, we the find:

$ \displaystyle U(12,4)=48<50$

And so we may conclude that:

$ \displaystyle U_{\max}=50$

We also find that:

$ \displaystyle \lambda=1$