Question

How do you maximize and minimize `f(x,y)=x-siny` constrained to `0<=x+y<=1`?

Answer 1

Write a Lagrange function with 2 multipliers and 2 slack variables.
Compute the partial derivatives.
Solve the system of equations.

Explanation:

For the constraint, `0<=x+y<=1`, written separately:

`x+y>=0` and `1-x - y>= 0`

We add two slack variables into two constraint functions:

`g_1(x,y,s) = x+y-s^2=0`

`g_2(x,y,t) = 1-x-y-t^2=0`

Note: squaring the slack variables assures that the constraint is enforced by disallowing negative values.

We can write the Lagrange function:

`\mathcalL(x,y,s,t,u,v) = f(x,y)+ug_1(x,y,s)+vg_2(x,y,t)`

`\mathcalL(x,y,s,t,u,v) = x-sin(y)+u(x+y-s^2)+v(1-x-y-t^2)`

`\mathcalL(x,y,s,t,u,v) = x-sin(y)+ux+uy-us^2+v-vx-vy-vt^2`

Compute the partial derivatives:

`(del(\mathcalL(x,y,s,t,u,v)))/(delx) = 1+u-v`

`(del(\mathcalL(x,y,s,t,u,v)))/(dely) = -cos(y)+u-v`

`(del(\mathcalL(x,y,s,t,u,v)))/(dels) = -2us`

`(del(\mathcalL(x,y,s,t,u,v)))/(delt) = -2vt`

`(del(\mathcalL(x,y,s,t,u,v)))/(delu) = x+y-s^2`

`(del(\mathcalL(x,y,s,t,u,v)))/(delv) = 1-x-y-t^2`

Set the partial derivatives equal to 0 and the solve as a system of nonlinear equations:

`0 = 1+u-v" [1]"`

`0 = -cos(y)+u-v" [2]"`

`0 = -2us" [3]"`

`0 = -2vt" [4]"`

`0 = x+y-s^2" [5]"`

`0 = 1-x-y-t^2" [6]"`

Please observe that the extrema are located at the points are where equations [3] and [4] are satisfied by `s = t= 0`; this makes the remaining equations become:

`0 = 1+u-v" [1]"`

`0 = -cos(y)+u-v" [2]"`

`0 = x+y" [5.1]"`

`0 = 1-x-y" [6.1]"`

Because u and v are not forced to be zero, we can subtract equation [2] from equation [1]

`0 = 1 +cos(y)`

`cos(y) = -1`

`y = cos^-1(-1)`

`y = pi`

Using equation [5.1] , we obtain the value for x:

`x = -pi`

This gives the function's minimum:

`f(-pi,pi) = -pi`

For the maximum, we have the condition where `u =s = t = 0`

`cos(y) = 1`

`y = 0`

Equation [6.2] gives us the value for x:

`x = 1`

`f(1,0) = 1`