Question Number 220869 by fantastic last updated on 20/May/25
$${Find}\:{the}\:{maximum}\:{value}\:{of}\:{x}^{\mathrm{2}} {y}^{\mathrm{3}} {z}^{\mathrm{4}} \:{subject}\:{to}\:{the}\:{condition}\:{x}+{y}+{z}=\mathrm{18} \\ $$
Commented by mr W last updated on 20/May/25
$${x},\:{y},\:{z}\:\in{N}\:\:\:{or}\:{Z}\:{or}\:{R}\:? \\ $$
Answered by Ghisom last updated on 20/May/25
![obviously x, y, z >0∧x<y<z let y=px∧z=qx∧p, q >0 x+y+z=18 ⇒ q=((18)/x)−p−1 x^2 y^3 z^4 =p^3 x^5 ((p+1)x−18)^4 (d/dp)[p^3 x^5 ((p+1)x−18)^4 ]=0 p^2 x^5 ((p+1)x−18)^3 ((7p+3)x−54)=0 ⇒ p=((54)/(7x))−(3/7)[∨p=((18)/x)−1 ⇒ q=0 rejected] q=((72)/(7x))−(4/7) x^2 y^3 z^4 =((6912x^2 (18−x)^7 )/(823543)) (d/dx)[((6912x^2 (18−x)^7 )/(823543))]=0 ((62208x(x−18)^6 (4−x))/(823543))=0 ⇒ x=4∧y=6∧z=8 x^2 y^3 x^4 =14155776](https://www.tinkutara.com/question/Q220907.png)
$$\mathrm{obviously}\:{x},\:{y},\:{z}\:>\mathrm{0}\wedge{x}<{y}<{z} \\ $$$$\mathrm{let}\:{y}={px}\wedge{z}={qx}\wedge{p},\:{q}\:>\mathrm{0} \\ $$$${x}+{y}+{z}=\mathrm{18}\:\Rightarrow\:{q}=\frac{\mathrm{18}}{{x}}−{p}−\mathrm{1} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{3}} {z}^{\mathrm{4}} ={p}^{\mathrm{3}} {x}^{\mathrm{5}} \left(\left({p}+\mathrm{1}\right){x}−\mathrm{18}\right)^{\mathrm{4}} \\ $$$$\frac{{d}}{{dp}}\left[{p}^{\mathrm{3}} {x}^{\mathrm{5}} \left(\left({p}+\mathrm{1}\right){x}−\mathrm{18}\right)^{\mathrm{4}} \right]=\mathrm{0} \\ $$$${p}^{\mathrm{2}} {x}^{\mathrm{5}} \left(\left({p}+\mathrm{1}\right){x}−\mathrm{18}\right)^{\mathrm{3}} \left(\left(\mathrm{7}{p}+\mathrm{3}\right){x}−\mathrm{54}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{p}=\frac{\mathrm{54}}{\mathrm{7}{x}}−\frac{\mathrm{3}}{\mathrm{7}}\left[\vee{p}=\frac{\mathrm{18}}{{x}}−\mathrm{1}\:\Rightarrow\:{q}=\mathrm{0}\:\mathrm{rejected}\right] \\ $$$${q}=\frac{\mathrm{72}}{\mathrm{7}{x}}−\frac{\mathrm{4}}{\mathrm{7}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{3}} {z}^{\mathrm{4}} =\frac{\mathrm{6912}{x}^{\mathrm{2}} \left(\mathrm{18}−{x}\right)^{\mathrm{7}} }{\mathrm{823543}} \\ $$$$\frac{{d}}{{dx}}\left[\frac{\mathrm{6912}{x}^{\mathrm{2}} \left(\mathrm{18}−{x}\right)^{\mathrm{7}} }{\mathrm{823543}}\right]=\mathrm{0} \\ $$$$\frac{\mathrm{62208}{x}\left({x}−\mathrm{18}\right)^{\mathrm{6}} \left(\mathrm{4}−{x}\right)}{\mathrm{823543}}=\mathrm{0} \\ $$$$\Rightarrow\:{x}=\mathrm{4}\wedge{y}=\mathrm{6}\wedge{z}=\mathrm{8} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{3}} {x}^{\mathrm{4}} =\mathrm{14155776} \\ $$
Commented by fantastic last updated on 20/May/25
$${OMG} \\ $$
Commented by Ghisom last updated on 20/May/25
$$\mathrm{what}?! \\ $$
Commented by fantastic last updated on 20/May/25
$$\:{sir}\:{in}\:{my}\:{book}\:{it}\:{used}\:{partial}\:{differention}\:!\:{and}\:{it}\:{is}\:{very}\:{lengthy} \\ $$$${i}\:{am}\:{very}\:{surprised} \\ $$