Question

If $3x+4y=5$, the greatest value of $x^2{y}^3$ is then:

• A. $\frac{3}{4}$
• B. $\frac{3}{8}$
• C. $\frac{3}{16}$
• D. $\frac{1}{16}$

Answer 1

The correct option is C $\frac{3}{16}$

Given, $3x+4y=5⟹3x=5-4y⟹x=\frac{5-4y}{3}$

substituting $x=\frac{5-4y}{3}$ in $x^2{y}^3$

$x^2{y}^3⟹(\frac{5-4y}{3}{)}^2{y}^3$

taking the derivative and equating to zero.

$⟹2\left(\frac{5-4y}{3}\right)y^3\left(\frac{-4}{3}\right)+3y^2(\frac{5-4y}{3}{)}^2=0$

$⟹y^2\frac{(5-4y)}{3}(-\frac{8}{3}y+(5-4y\left)\right)=0$

$⟹y^2=0$ and $(5-4y)=0$ and $(-\frac{20}{3}y+5)=0$

$⟹y=0$ and $y=\frac{5}{4}$ and $y=\frac{3}{4}$

Substituting $y=0⟹x=\frac{5-4\left(0\right)}{3}=\frac{5}{3}$

Substituting $y=\frac{5}{4}⟹x=\frac{5-4\left(5/4\right)}{3}=0$

Substituting $y=\frac{3}{4}⟹x=\frac{5-4\left(3/4\right)}{3}=\frac{2}{3}$

therefore, substituting $y=0$ and $x=\frac{5}{3}⟹x^2{y}^3=0$

substituting $y=\frac{5}{4}$ and $x=0⟹x^2{y}^3=0$

substituting $y=\frac{3}{4}$ and $x=\frac{2}{3}⟹x^2{y}^3=\left(\frac{2}{3}{)}^2\right(\frac{3}{4}{)}^3=\frac{3}{16}$

therefore, the greatest value is $\frac{3}{16}$