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  1. Pessimist Singularitarian
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    #11
    Quote Originally Posted by mrjericho1991 View Post
    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?

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    #12
    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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