for-x-y-z-gt-0-find-the-maximum-of-x-m-y-n-z-k-subject-to-ax-by-cz-d-

Question Number 220958 by mr W last updated on 21/May/25

for x, y, z >0 find the maximum of x^m y^n z^k subject to ax+by+cz=d.

$${for}\:{x},\:{y},\:{z}\:>\mathrm{0}\:{find}\:{the}\:{maximum}\:{of} \\ $$$${x}^{{m}} {y}^{{n}} {z}^{{k}} \:{subject}\:{to}\:{ax}+{by}+{cz}={d}. \\ $$

Commented by mr W last updated on 21/May/25

$${a}\:{general}\:{case}\:{to}\:{Q}\mathrm{220869} \\ $$

Answered by mr W last updated on 21/May/25

maximum of x^m y^n z^k ⇔ maximum of ln (x^m y^n z^k ) Φ=ln (x^m y^n z^k )=mln x+nln y+kln z F(x,y,z)=mln x+nln y+kln z+λ(ax+by+cz−d) (∂F/∂x)=(m/x)+λa=0 ⇒x=−(m/(aλ)) (∂F/∂y)=(n/y)+λb=0 ⇒y=−(n/(bλ)) (∂F/∂z)=(k/z)+λc=0 ⇒z=−(k/(cλ)) −a×(m/(aλ))−b×(n/(bλ))−c×(k/(cλ))=d ⇒(1/λ)=−(d/(m+n+k)) ⇒x=(dm/((m+n+k)a)) ⇒y=(dn/((m+n+k)b)) ⇒z=(dk/((m+n+k)c)) ⇒(x^m y^n z^k )_(max) =((d/(m+n+k)))^(m+n+k) ((m/a))^m ((n/b))^n ((k/c))^k example: m=2, n=3, k=4, a=b=c=1, d=18 (x^2 y^3 z^4 )_(max) =(((18)/(2+3+4)))^(2+3+4) 2^2 3^3 4^4 =14 155 776

$${maximum}\:{of}\:{x}^{{m}} {y}^{{n}} {z}^{{k}} \: \\ $$$$\Leftrightarrow\:{maximum}\:{of}\:\mathrm{ln}\:\left({x}^{{m}} {y}^{{n}} {z}^{{k}} \right) \\ $$$$\Phi=\mathrm{ln}\:\left({x}^{{m}} {y}^{{n}} {z}^{{k}} \right)={m}\mathrm{ln}\:{x}+{n}\mathrm{ln}\:{y}+{k}\mathrm{ln}\:{z} \\ $$$${F}\left({x},{y},{z}\right)={m}\mathrm{ln}\:{x}+{n}\mathrm{ln}\:{y}+{k}\mathrm{ln}\:{z}+\lambda\left({ax}+{by}+{cz}−{d}\right) \\ $$$$\frac{\partial{F}}{\partial{x}}=\frac{{m}}{{x}}+\lambda{a}=\mathrm{0}\:\Rightarrow{x}=−\frac{{m}}{{a}\lambda} \\ $$$$\frac{\partial{F}}{\partial{y}}=\frac{{n}}{{y}}+\lambda{b}=\mathrm{0}\:\Rightarrow{y}=−\frac{{n}}{{b}\lambda} \\ $$$$\frac{\partial{F}}{\partial{z}}=\frac{{k}}{{z}}+\lambda{c}=\mathrm{0}\:\Rightarrow{z}=−\frac{{k}}{{c}\lambda} \\ $$$$−{a}×\frac{{m}}{{a}\lambda}−{b}×\frac{{n}}{{b}\lambda}−{c}×\frac{{k}}{{c}\lambda}={d} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda}=−\frac{{d}}{{m}+{n}+{k}} \\ $$$$\Rightarrow{x}=\frac{{dm}}{\left({m}+{n}+{k}\right){a}} \\ $$$$\Rightarrow{y}=\frac{{dn}}{\left({m}+{n}+{k}\right){b}} \\ $$$$\Rightarrow{z}=\frac{{dk}}{\left({m}+{n}+{k}\right){c}} \\ $$$$\Rightarrow\left({x}^{{m}} {y}^{{n}} {z}^{{k}} \right)_{{max}} \\ $$$$\:\:\:\:=\left(\frac{{d}}{{m}+{n}+{k}}\right)^{{m}+{n}+{k}} \left(\frac{{m}}{{a}}\right)^{{m}} \left(\frac{{n}}{{b}}\right)^{{n}} \left(\frac{{k}}{{c}}\right)^{{k}} \\ $$$${example}:\: \\ $$$${m}=\mathrm{2},\:{n}=\mathrm{3},\:{k}=\mathrm{4},\:{a}={b}={c}=\mathrm{1},\:{d}=\mathrm{18} \\ $$$$\left({x}^{\mathrm{2}} {y}^{\mathrm{3}} {z}^{\mathrm{4}} \right)_{{max}} =\left(\frac{\mathrm{18}}{\mathrm{2}+\mathrm{3}+\mathrm{4}}\right)^{\mathrm{2}+\mathrm{3}+\mathrm{4}} \mathrm{2}^{\mathrm{2}} \mathrm{3}^{\mathrm{3}} \mathrm{4}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{14}\:\mathrm{155}\:\mathrm{776} \\ $$

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