Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 141164 by iloveisrael last updated on 16/May/21

Find maximum value of the product xy(72−3x−4y) for positive value of x & y.

$$\:\:\:\:{Find}\:{maximum}\:{value}\: \\ $$ $$\:\:\:{of}\:{the}\:{product}\:{xy}\left(\mathrm{72}−\mathrm{3}{x}−\mathrm{4}{y}\right) \\ $$ $$\:\:\:\:{for}\:{positive}\:{value}\:{of}\:{x}\:\&\:{y}. \\ $$

Commented bymitica last updated on 16/May/21

max=1152 for x=8,y=6

$${max}=\mathrm{1152}\:{for}\:{x}=\mathrm{8},{y}=\mathrm{6} \\ $$

Answered by EDWIN88 last updated on 16/May/21

The terms of the product x,y & 72−3x−4y do not have a constant sum. However if we adjust the product and write it as (1/(12))(3x)(4y)(72−3x−4y) then ignoring the multiplier (1/(12)) for a moment we see that the factors 3x,4y & 72−3x−4y have a constant 72. Thus we seek value of x & y satisfying 3x=4y=72−3x−4y given the unique values { ((x=8)),((y=6)) :} Thus the max value of xy(72−3x−4y) is 8×6×24 = 1,152 ⋇

$$\mathrm{The}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{product}\:\mathrm{x},\mathrm{y}\:\&\:\mathrm{72}−\mathrm{3x}−\mathrm{4y} \\ $$ $$\mathrm{do}\:\mathrm{not}\:\mathrm{have}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{sum}.\:\mathrm{However}\:\mathrm{if}\:\mathrm{we} \\ $$ $$\mathrm{adjust}\:\mathrm{the}\:\mathrm{product}\:\mathrm{and}\:\mathrm{write}\:\mathrm{it}\:\mathrm{as}\: \\ $$ $$\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{3x}\right)\left(\mathrm{4y}\right)\left(\mathrm{72}−\mathrm{3x}−\mathrm{4y}\right)\: \\ $$ $$\:\mathrm{then}\:\mathrm{ignoring}\:\mathrm{the}\:\mathrm{multiplier}\:\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{for}\:\mathrm{a}\:\mathrm{moment} \\ $$ $$\mathrm{we}\:\mathrm{see}\:\mathrm{that}\:\mathrm{the}\:\mathrm{factors}\:\mathrm{3x},\mathrm{4y}\:\&\:\mathrm{72}−\mathrm{3x}−\mathrm{4y} \\ $$ $$\:\mathrm{have}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{72}.\:\mathrm{Thus}\:\mathrm{we}\:\mathrm{seek}\:\mathrm{value}\:\mathrm{of} \\ $$ $$\:\mathrm{x}\:\&\:\mathrm{y}\:\mathrm{satisfying}\: \\ $$ $$\:\:\:\:\:\:\:\mathrm{3x}=\mathrm{4y}=\mathrm{72}−\mathrm{3x}−\mathrm{4y} \\ $$ $$\:\mathrm{given}\:\mathrm{the}\:\mathrm{unique}\:\mathrm{values}\:\begin{cases}{\mathrm{x}=\mathrm{8}}\\{\mathrm{y}=\mathrm{6}}\end{cases} \\ $$ $$\:\mathrm{Thus}\:\mathrm{the}\:\mathrm{max}\:\mathrm{value}\:\mathrm{of}\:\mathrm{xy}\left(\mathrm{72}−\mathrm{3x}−\mathrm{4y}\right)\:\mathrm{is} \\ $$ $$\:\:\:\:\:\:\:\mathrm{8}×\mathrm{6}×\mathrm{24}\:=\:\mathrm{1},\mathrm{152}\:\divideontimes \\ $$

Commented byiloveisrael last updated on 16/May/21

Yes....

$$\mathrm{Y}{es}.... \\ $$

Answered by mitica last updated on 16/May/21

if 72−3x−4y≤0⇒xy(72−3x−4y)≤0 if 72−3x−4y>0⇒(1/(12))∙3x∙4y∙(72−3x−4y)≤^(am−gm) (1/(12))(3x+4y−72−3x−4y)^3 ∙(1/(27))=1152 = for 3x=4y=72−3x−4y⇒x=8,y=6 ⇒max{xy(72−3x−4y)}=1152

$${if}\:\mathrm{72}−\mathrm{3}{x}−\mathrm{4}{y}\leqslant\mathrm{0}\Rightarrow{xy}\left(\mathrm{72}−\mathrm{3}{x}−\mathrm{4}{y}\right)\leqslant\mathrm{0} \\ $$ $${if}\:\mathrm{72}−\mathrm{3}{x}−\mathrm{4}{y}>\mathrm{0}\Rightarrow\frac{\mathrm{1}}{\mathrm{12}}\centerdot\mathrm{3}{x}\centerdot\mathrm{4}{y}\centerdot\left(\mathrm{72}−\mathrm{3}{x}−\mathrm{4}{y}\right)\overset{{am}−{gm}} {\leqslant}\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{3}{x}+\mathrm{4}{y}−\mathrm{72}−\mathrm{3}{x}−\mathrm{4}{y}\right)^{\mathrm{3}} \centerdot\frac{\mathrm{1}}{\mathrm{27}}=\mathrm{1152} \\ $$ $$=\:{for}\:\mathrm{3}{x}=\mathrm{4}{y}=\mathrm{72}−\mathrm{3}{x}−\mathrm{4}{y}\Rightarrow{x}=\mathrm{8},{y}=\mathrm{6} \\ $$ $$\Rightarrow{max}\left\{{xy}\left(\mathrm{72}−\mathrm{3}{x}−\mathrm{4}{y}\right)\right\}=\mathrm{1152} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com