Maximize P = xyz with x + y = 36 and y + z = 36, and x, y, and z ≥ 0.

Question

surjithayer · Answer

any idea?

surjithayer · Answer

@surjithayer wrote:

let us check if we can reach to the solution. x+y=36 x=36-y y+z=36 z=36-y so x=z $$p=xyz=y(36-y)(36-y)=y(36-y)^2$$ $$\frac{ dp }{ dy }$$$$=(36-y)^2+y[2(36-y)(-1)]$$$$=(36-y)[36-y-2y]$$$$=(36-y)(36-3y)$$ $$\frac{ dp }{ dy }=0,gives (36-y)(36-3y)=0,y=36,12$$ $$\frac{ d^2p }{ dy^2}=(36-y)(-3)+(-1)(36-3y)$$$$=-108+3y-36+3y=6y-144$$ at y=36 $$\frac{ d^2p }{ dy^2 }=6(36)-144=216-144=72>0$$ p is minimum at y=36 when y=12 $$\frac{ d^2p }{ dy^2 }=6(12)-144=72-144=-72<0$$ p is maximum at y=12 x=z=36-12=24 max. value of p=xyz=24 *24*12=576*12=6912