Given: #f(x,y)=x^2/y^3+xy^2#
A little better form:
#f(x,y) = x^2y^-3+xy^2#
We need to work on the constraint functions.
#g_1(x,y,s_1) = x+3xy-s_1^2=0#
#g_2(x,y,s_2) = 4-x-3xy-s_2^2=0#
NOTE: #s_1 and s_2# are called "slack variables" they allow us to take up the slack for the inequality.
We have two constraint functions, therefore we have two Lagrange multipliers and the Lagrange function is:
#L(x,y,lambda_1,lambda_2,s_1,s_2) = x^2y^-3+xy^2 +lambda_1x+3lambda_1xy-lambda_1s_1^2+4lambda_2-lambda_2x-3lambda_2xy-lambda_2s_2^2#
The six partial derivatives are:
#(delL(x,y,lambda_1,lambda_2,s_1,s_2))/(delx) = 2xy^-3+y^2+lambda_1(1+3y) -lambda_2(1+3y)#
#(delL(x,y,lambda_1,lambda_2,s_1,s_2))/(dely) = -6xy^-4+2xy+3x(lambda_1-lambda_2)#
#(delL(x,y,lambda_1,lambda_2,s_1,s_2))/(dellambda_1) = x+3xy-s_1^2#
#(delL(x,y,lambda_1,lambda_2,s_1,s_2))/(dellambda_2) = 4-x-3xy-s_2^2#
#(delL(x,y,lambda_1,lambda_2,s_1,s_2))/(s_1) = -2lambda_1s_1#
#(delL(x,y,lambda_1,lambda_2,s_1,s_2))/(s_2) = -2lambda_2s_2#
Set them equal to zero and solve them as a system of equations:
#0 = 2xy^-3+y^2+lambda_1(1+3y) -lambda_2(1+3y)#
#0 = -6xy^-4+2xy+3x(lambda_1-lambda_2)#
#0 = x+3xy-s_1^2#
#0 = 4-x-3xy-s_2^2#
#0 = -2lambda_1s_1#
#0 = -2lambda_2s_2#
I substituted #u,v,w, and z# for #lambda_1, lambda_2, s_1 and s_2#, respectively and gave the equations to WolframAlpha. It came up with:
#x~~0.754146, y~~1.43467#
