Question

On the interval \[\left[ {0,1} \right]\], the function \[{x^{25}}{\left( {1 - x} \right)^{75}}\] takes its maximum value at the point
A. \[0\]
B. \[\dfrac{1}{4}\]
C. \[\dfrac{1}{2}\]
D. \[\dfrac{1}{3}\]

Answer 1

Hint: In this question, we need to find the value of x in the given interval at which the function \[{x^{25}}{\left( {1 - x} \right)^{75}}\] takes its maximin value. For that, we differentiate the given function then we put \[f^{'}\left( x \right) = 0\] then we put all values in f(x) to get the desired result.

Formula Used:
1. The product rule of differentiation is \[\dfrac{d}{{dx}}\left( {u \cdot v} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]
2. The chain rule of differentiation is \[\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f^{'}\left( {g\left( x \right)} \right) \cdot g^{'}\left( x \right)\]
3. \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\] where n is the real number.

Complete step-by-step solution:
We are given that \[{x^{25}}{\left( {1 - x} \right)^{75}},x \in \left[ {0,1} \right]\]
Now we assume that \[f\left( x \right) = {x^{25}}{\left( {1 - x} \right)^{75}},x \in \left[ {0,1} \right]\]
Now by differentiation with respect to x, we get
     \[f^{'}\left( x \right) = \dfrac{d}{{dx}}\left( {{x^{25}}{{\left( {1 - x} \right)}^{75}}} \right)...\left( 1 \right)\]
Now by applying the product rule of differentiation, we get
     \[f^{'}\left( x \right) = {x^{25}}\dfrac{d}{{dx}}{\left( {1 - x} \right)^{75}} + {\left( {1 - x} \right)^{75}}\dfrac{d}{{dx}}\left( {{x^{25}}} \right)\]
We know that \[\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}\]and \[\dfrac{d}{{dx}}f\left( {g\left( x \right)} \right) = f^{'}\left( {g\left( x \right)} \right) \cdot g^{'}\left( x \right)\]
\[
  f^{'}\left( x \right) = {x^{25}}\left( {75{{\left( {1 - x} \right)}^{75 - 1}}\left( { - 1} \right)} \right) + {\left( {1 - x} \right)^{75}}\left( {25{x^{25 - 1}}} \right) \\
   = x^{25}\left( { - 75{{\left( {1 - x} \right)}^{74}}} \right) + 25{\left( {1 - x} \right)^{75}}{x^{24}} \\
   = - 75{x^{25}}{\left( {1 - x} \right)^{74}} + 25{\left( {1 - x} \right)^{75}}{x^{24}}
 \]
Now we take common terms out and simplify them, we get
\[
f^{'}\left ( x \right )=25\left ( 1-x \right )^{74}x^{24}\left [ -3x+\left ( 1-x \right ) \right ]\\
=25\left ( 1-x \right )^{74}x^{24}\left ( -3x+1-x \right )\\
=25\left ( 1-x \right )^{74}x^{24}\left ( -4x+1 \right )\\
=25\left ( 1-x \right )^{74}x^{24}\left ( 1-4x \right )
\]
Now we substitute \[f^{'}\left( x \right) = 0\], we get
\[25{\left( {1 - x} \right)^{74}}{x^{24}}\left( {1 - 4x} \right) = 0\]
Now when we substitute \[{x^{24}} = 0\], we get
\[x = 0\]
Now when we substitute \[{\left( {1 - x} \right)^{74}} = 0\], we get
\[
  1 - x = 0 \\
  1 = x \\
  x = 1 \\
 \]
Now when we substitute \[\left( {1 - 4x} \right) = 0\], we get
\[
  1 = 4x \\
  \dfrac{1}{4} = x \\
  x = \dfrac{1}{4} \\
 \]
Therefore, the value of x is \[x = 0,1,\dfrac{1}{4}\]
Now we substitute these values in the given function, and we get
Now when \[x = 0\]
\[
  f\left( 0 \right) = {0^{25}}{\left( {1 - 0} \right)^{75}} \\
   = 0{\left( 1 \right)^{75}} \\
   = 0 \\
 \]
Now when \[x = 1\]
\[
  f\left( 1 \right) = {1^{25}}{\left( {1 - 1} \right)^{75}} \\
   = 1{\left( 0 \right)^{75}} \\
   = 0 \\
 \]
Now when \[x = \dfrac{1}{4}\]
\[
  f\left( {\dfrac{1}{4}} \right) = {\left( {\dfrac{1}{4}} \right)^{25}}{\left( {1 - \dfrac{1}{4}} \right)^{75}} \\
   = {\left( {\dfrac{1}{4}} \right)^{25}}{\left( {\dfrac{{4 - 1}}{4}} \right)^{75}} \\
   = {\left( {\dfrac{1}{4}} \right)^{25}}{\left( {\dfrac{3}{4}} \right)^{75}} \\
   > 0 \\
 \]
Therefore, the maximum value of the given function \[f\left( x \right) = {x^{25}}{\left( {1 - x} \right)^{75}}\] is \[\dfrac{1}{4}\].

Hence, option (B) is correct


Note: Students can make mistakes when calculating the derivatives of a given function because the function involves large numbers. Students must be sure that the discovered solution falls between (0,1). Students can also do this solution by taking the second derivative of the function.