Question

Find the greatest value of
${x}^{3} {y}^{4} \: when \: 2x + 3y = 7 \: and \: x \geqslant 0 \: and \: y \geqslant 0$
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x,y \: \geqslant \: 0$

and

$\rm :\longmapsto\:2x + 3y = 7$

can be rewritten as

$\rm :\longmapsto\:3\bigg[\dfrac{2x}{3}\bigg] + 4\bigg[\dfrac{3y}{4}\bigg]$

can be further rewritten as

$\rm :\longmapsto\:\bigg[\dfrac{2x}{3}\bigg] + \bigg[\dfrac{2x}{3}\bigg] + \bigg[\dfrac{2x}{3}\bigg] + \bigg[\dfrac{3y}{4}\bigg] + \bigg[\dfrac{3y}{4}\bigg] + \bigg[\dfrac{3y}{4}\bigg] + \bigg[\dfrac{3y}{4}\bigg] = 7$

So, greatest value of

$\rm :\longmapsto\:\bigg[\dfrac{2x}{3}\bigg]\bigg[\dfrac{2x}{3}\bigg]\bigg[\dfrac{2x}{3}\bigg]\bigg[\dfrac{3y}{4}\bigg]\bigg[\dfrac{3y}{4}\bigg]\bigg[\dfrac{3y}{4}\bigg]\bigg[\dfrac{3y}{4}\bigg]= {\bigg[\dfrac{7}{7} \bigg]}^{7}$

$\rm :\longmapsto\:\dfrac{ {2}^{3} \: {3}^{4} }{ {3}^{3} \: {4}^{4} } {x}^{3} {y}^{4} = 1$

$\rm :\longmapsto\:\dfrac{3}{32 } {x}^{3} {y}^{4} = 1$

$\bf\implies \: {x}^{3} {y}^{4} = \dfrac{32}{3}$

$\bf\implies \: Greatest \: value \: of \: {x}^{3} {y}^{4} = \dfrac{32}{3}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Basic Concept Used

Maxima and Minima

$\rm :\longmapsto\:a_1,a_2, - - ,a_n \: are \: n \: positive \: integers \: such \: that \:$

$\rm :\longmapsto\:a_1 + a_2 + - - - + a_n = k \: (constant) \: then$

$\rm :\longmapsto\:a_1a_2 - - - a_n \: is \: greatest \: when \: a_1 = a_2 = - - = a_n$

$\rm :\longmapsto\:greatest \: valu e \: of \: a_1a_2 - - - a_n = {\bigg[\dfrac{k}{n} \bigg]}^{n}$