Question

Find the maximum value of $x^3{(4a-x)}^5$ if $x$ is positive and less than $4a$; and the maximum value of $x^{\frac{1}{2}}{(1-x)}^{\frac{1}{3}}$ when $x$ is a proper fraction.

Answer 1

Let $P=x^3{(4a-x)}^5$

$P=3^3\frac{x^3}{3^3}\times 5^5\frac{{(4a-x)}^5}{5^5}$

As,$3\frac{x}{3}\times 5\frac{(4a-x)}{5}$

$=x+4a-x$

$=4a$ which is a constant

So , $\frac{x^3}{3^3}\times \frac{{(4a-x)}^5}{5^5}$ will be maximum if ,

$\frac{x}{3}=\frac{4a-x}{5}$

$5x=12a-3x$

$8x=12a$

$x=\frac{3a}{2}$

$P$ will be maximum if ${\left(\frac{3a}{2}\right)}^3\times {(4a-\frac{3a}{2})}^5$

$=\frac{27a^3}{8}\frac{5^5{a}^5}{2^5}=\frac{84375}{256}$

Now $P=x^{\frac{1}{2}}{(1-x)}^{\frac{1}{3}}$

$P^6=x^3{(1-x)}^2$

$=3^3{\left(\frac{x}{3}\right)}^3{2}^2{\left(\frac{(1-x)}{2}\right)}^2$

$P$ wil be maximum when ${\left(\frac{x}{3}\right)}^3{\left(\frac{(1-x)}{2}\right)}^2$ will be maximum

Sum of $3\frac{x}{3}+2\frac{1-x}{2}=1$ which is constant

${\left(\frac{x}{3}\right)}^3{\left(\frac{(1-x)}{2}\right)}^2$ will be maximum if ,

$\frac{x}{3}=\frac{1-x}{2}$

$2x=3-3x$

$x=\frac{3}{5}$

$P^6$ will be maximum if ${\left(\frac{3}{5}\right)}^3{(1-\frac{3}{5})}^2$

$P$ will be maximum ${\left(\frac{3}{5}\right)}^{\frac{1}{2}}{\left(\frac{2}{5}\right)}^{\frac{1}{3}}$