Maximizing and Minimizing Multivariable Equations

  • Thread starter Thread starter Pengwuino
  • Start date Start date
  • Tags Tags
    3d
Pengwuino
Gold Member
Messages
5,112
Reaction score
20
Hey guys I am very confused here.. I have no idea how to do this! I need to find the max/mins of this equation and I get lost after finding the partials.

f(x,y) = xy(1 - x - y) \\
f_x (x,y) = y - 2xy - y^2 \\
f_y (x,y) = x - x^2 - 2xy \\

I know I'm suppose to kinda do what you do with single variables... but I am getting lost with these multivariables... any help would be very much appreciated :)
 
Physics news on Phys.org
To find relative extrema set the gradient of the function to zero. Once you've found candidates, you'll need to determine whether the points are minima, maxima or saddle points.
 
that doesn't seem like how we were being taught in the book. It seemed like they were doing it all algebraicly
 
more simply: set f_{x}=f_{y}=0 and solve the system for x and y
 
IOW, like benorin said, set the derivatives equal to zero.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top