Question

Find the maximum value of ${x^m}{y^m}$ where m. n>0 such that $x + y = a$.

Answer 1

The maximum value of a function is the place where a function reaches its highest point on a graph.
If the sum of positive variables is constant their product will be maximum, when all the variables are equal to each other.
If the product of positive variables is constant there sum will be minimum, when all variables are equal to each other.
Form values for A.M and G.M are: -
$A.M =$ arithmetic means ${a_1} + {a_2} + .... + an$
$G.M. =$ Geometric means $= {\sqrt[n]{{{a_1}{a_2}a}}_3}....an$
Where ${a_1},{a_2}...an$ are n terms
If A.M of set of positive number $\geqslant$ G.M of some set of positive numbers
To find the maximum of a function we need to put its derivative equals to zero.

Complete step-by-step answer:
Let $z = {x^m}{y^n} - \,\,\,\,\,eqn(1)$
Given $x + y = a$
We have $y = a - x$
Substitute value of y in $eqn(i)$, we get
$z = {x^m}{(a - x)^n}\,\,\,\,\,eqn(2)$
To find maximum we need to derivate
Equation (2) with respect to x and put it as zero
$i.e\dfrac{{d > }}{{dx}} = 0$
Derivation equation (2) with respect to x, using product rule
$\dfrac{{d > }}{{dx}} = {x^m}n{(a - x)^{n - 1}}\dfrac{d}{{dx}}{(a - x)^n} + {(a - x)^n}m{x^{m - 1}} = 0$
$\left[ {\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}} \right]$
$\dfrac{d}{{dx}} = - {x^m}n{(a - x)^{n - 1}} + m{(a - x)^n}{x^{m - 1}}$
$\left[ {\dfrac{d}{{dx}}(a - x) = - 1} \right]$
$\Rightarrow {(a - x)^n}\,m{\dfrac{x}{x}^m} = {x^m}n\dfrac{{{{(a - x)}^n}}}{{(a - x}}$
${(a - x)^n}\,$and ${x^m}$ will be eliminated from both sides
$\Rightarrow \dfrac{m}{x} = \dfrac{n}{{a - x}}$
Cross multiplying we get
$m \times (a - x) = n \times x$
$ma - mx = nx$
$ma - nx + mx$
$ma - (n + m)x$
$x = \dfrac{{ma}}{{m + n}}\,\,\,\,eqn(3)$
Using $\,eqn(3)$ in $\,eqn(2)$, we get
$z\max = {\left( {\dfrac{{ma}}{{m + n}}} \right)^m}{(a - \dfrac{{ma}}{{m + n}})^n}$
Now, multiply a to numerator to simplify
$z\max = {\left( {\dfrac{{ma}}{{m + n}}} \right)^m}{(\dfrac{{ma + na - ma}}{{m + n}})^n}$
Now, we’ll cancel out common with opposite sign
$z\max = {\left( {\dfrac{{ma}}{{m + n}}} \right)^m}{(\dfrac{{na}}{{m + n}})^n}$
Multiply,
$z\max = \dfrac{{{{(ma)}^m}{{(na)}^n}}}{{{{(m + n)}^m}{{(m + n)}^n}}}$
$z\max = \dfrac{{{m^m}{n^n}{a^n}}}{{{{(m + n)}^m}{{(m + n)}^n}}}$
$z\max = \dfrac{{{m^n}{n^n}{a^{m + n}}}}{{{{(m + n)}^{m + n}}}}\,\,\,\,$ {Same base power will add}

Note: It can be done with another way as well as
$A.M \geqslant G.M$
$x + y = a \Rightarrow m\left( {\dfrac{x}{m}} \right) + n\left( {\dfrac{y}{x}} \right) = a$
We know, $A.M \geqslant G.M$
\Rightarrow \dfrac{{m\left( {\dfrac{x}{m}} \right) + n\left( {\dfrac{y}{x}} \right)}}{{m + n}}$$$$ \geqslant \sqrt[{m + n}]{{{{\left( {\dfrac{x}{m}} \right)}^m}{{\left( {\dfrac{y}{m}} \right)}^n}}}
${x^m}{y^n} \leqslant \dfrac{{{a^{m + n}}{m^m}.{n^n}}}{{{{(m + n)}^{m + n}}}}$
The maximum value is $\dfrac{{{a^{m + n}}{m^m}{n^n}}}{{{{(m + n)}^{m + n}}}}$