Exercise 2115Calculus › Multivariable calculus › Optimisation in several variablesM1Added 3 Nov 2026

Two equality constraints

#Lagrange#multiple constraints#KKT

Statement

Maximize $f(x,y,z)=xyz$ under $x+y+z=3$ and $x=y$ .

Hint

Substitute $y=x$ first, then Lagrange on two variables.

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Substitute. With $y=x$ : $f(x,z)=x^2z$ , constraint $2x+z=3$ .

Lagrange. $\partial_x f=2xz=\lambda\cdot 2$ , $\partial_z f=x^2=\lambda$ . So $xz=x^2/2$ , i.e. $z=x/2$ . Then $2x+x/2=3$ , $x=6/5$ , $z=3/5$ .

Value. $f=(6/5)^2\cdot3/5=108/125$ .

Conclusion. Maximum $108/125$ at $(6/5,6/5,3/5)$ .