Question

When z=f(x,y)=2x+6y+1 is constrained by x²+y²=40, what become the minimum and maximum values?

Answer 1

given z=f(x,y)=2x+6y+1

x²+y²=40
=>x²+y²-40=0

let g(x,y)=x²+y²-40

using the method of lagrange multipliers
∇f=m∇g
=>〈2,6〉=m〈2x,2y〉
=>〈2,6〉=〈2xm,2ym〉

2xm=2,2ym=6
=>m=1_/x,m=3_/y
=>1_/x=3_/y
=>y=3x

x²+y²=40,y=3x
=>x²+(3x)²=40
=>x²+9x²=40
=>10x²=40
=>x²=4
=>x=-2,x=2

y=3x,x=-2
=>y=3(-2)
=>y=-6

y=3x,x=2
=>y=3(2)
=>y=6

(-2,-6),(2,6) are candidate points

f(-2,-6)=2(-2)+6(-6)+1
=>f(-2,-6)=-4-36+1
=>f(-2,-6)=-39

f(2,6)=2(2)+6(6)+1
=>f(2,6)=4+36+1
=>f(2,6)=41

minimum value is -39
maximum value is 41