# Thread: Maximizing profit and utility functions

1. Originally Posted by mrjericho1991
without constraint. Yes I know but tell me what values do I need to input for partial derivatives?
You need to equate the first partials to zero, and then solve the resulting system. We have:

$\displaystyle P(x,y)=100x+80y+2xy-x^2-2y^2-5000$

And so we compute the first partials, and equation them to zero to obtain the system:

$\displaystyle 100+2y-2x=0$

$\displaystyle 80+2x-4y=0$

So, what critical point do you get from solving the above linear system?

2. 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:

$\displaystyle \lambda=1$

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