Question

If \(a\) and \(b\) are positive numbers, find the maximum value of \(f\left( x \right) = {x^a}{\left( {1 - x} \right)^b},\,\,0 \le x \le 1\).

Short answer

The absolute maximum value of the function \(f\left( x \right) = {x^a}{\left( {1 - x} \right)^b},0 \le x \le 1\) is \(f\left( {\frac{a}{{a + b}}} \right) = \frac{{{a^a}{b^b}}}{{{{\left( {a + b} \right)}^{a + b}}}}\).

Step-by-step solution

The Closed Interval Method

There are three steps to find the absolute maximum and minimum values of a continuous function\(f\)in any closed interval\(\left( {a,b} \right)\):

1. Byfinding critical numbers, find the values of\(f\)at the critical numbers in\(\left( {a,b} \right)\).
2. Find the value of the function at eachendpoint of the interval.
3. Identify the largest value and smallest value obtained from step 1 and 2, wherelargest value will be absolute maximum value and the smallest value will be absolute minimum value.

Find the critical points

Consider the given function\(f\left( x \right) = {x^a}{\left( {1 - x} \right)^b}\). Differentiate the function w.r.t\(x\)by using product rule of differentiation.

\(\begin{aligned}{c}\frac{d}{{dx}}\left( {f\left( x \right)} \right) &= \frac{d}{{dx}}\left( {{x^a}{{\left( {1 - x} \right)}^b}} \right)\\f'\left( x \right) &= {x^a}\frac{d}{{dx}}\left( {{{\left( {1 - x} \right)}^b}} \right) + {\left( {1 - x} \right)^b}\frac{d}{{dx}}\left( {{x^a}} \right)\\ &= {x^a} \cdot b{\left( {1 - x} \right)^{b - 1}} + {\left( {1 - x} \right)^b} \cdot a{x^{a - 1}}\\ &= {x^{a - 1}}{\left( {1 - x} \right)^{b - 1}}\left( {x \cdot b\left( { - 1} \right) + \left( {1 - x} \right)a} \right)\\ &= {x^{a - 1}}{\left( {1 - x} \right)^{b - 1}}\left( {a - ax - bx} \right)\end{aligned}\)

Substitute\(f'\left( x \right) = 0\)and solve for\(x\)to get the critical points.

\(\begin{aligned}{c}{x^{a - 1}}{\left( {1 - x} \right)^{b - 1}}\left( {a - ax - bx} \right) &= 0\\x &= 0,1,\frac{a}{{a + b}}\end{aligned}\)

Find value of \(f\)at critical points in \(\left( {0,1} \right)\)

At\(x = 0\),

\(\begin{aligned}{c}f\left( 0 \right) &= {0^a}{\left( {1 - 0} \right)^b}\\ &= 0\end{aligned}\)

At\(x = \frac{a}{{a + b}}\),

\(\begin{aligned}{c}f\left( {\frac{a}{{a + b}}} \right) &= {\left( {\frac{a}{{a + b}}} \right)^a}{\left( {1 - \frac{a}{{a + b}}} \right)^b}\\ &= {\left( {\frac{a}{{a + b}}} \right)^a}{\left( {\frac{{a + b - a}}{{a + b}}} \right)^b}\\ &= {\left( {\frac{a}{{a + b}}} \right)^a}{\left( {\frac{b}{{a + b}}} \right)^b}\\ &= \frac{{{a^a}{b^b}}}{{{{\left( {a + b} \right)}^{a + b}}}}\end{aligned}\)

Do not find at \(x = 1\), because it does not include in the interval \(\left( {0,1} \right)\).

Find value of \(f\) at endpoints of \(\left( {0,1} \right)\)

At\(x = 1\)

\(\begin{aligned}{c}f\left( 0 \right) &= {1^a}{\left( {1 - 1} \right)^b}\\ &= 0\end{aligned}\)

Observe the largest and smallest value of \(f\)

It can be observed that, at\(x = 0,1\),\(f\left( x \right) = 0\)which is the minimum value that will not be the absolute maximum value, and at\(x = \frac{a}{{a + b}}\),\(f\left( x \right) = \frac{{{a^a}{b^b}}}{{{{\left( {a + b} \right)}^{a + b}}}}\)or\(f\left( {\frac{a}{{a + b}}} \right) = \frac{{{a^a}{b^b}}}{{{{\left( {a + b} \right)}^{a + b}}}}\)is the maximum value among all the values, that will be the absolute maximum value.

Thus, the absolute maximum value of the function \(f\left( x \right) = {x^a}{\left( {1 - x} \right)^b}\) is\(f\left( {\frac{a}{{a + b}}} \right) = \frac{{{a^a}{b^b}}}{{{{\left( {a + b} \right)}^{a + b}}}}\).