Question

Find two positive integers x and y such that $ x + y = 60 $ and $ x{y^3} $ is maximum.

Answer 1

Hint: In this question, we need to evaluate the value of ‘x’ and ‘y’ such that the sum of the numbers be 60 and $ x{y^3} $ should be maximum. For this, we will differentiate the later function with respect to a variable and equate the result to zero to get the result.

Complete step-by-step answer:
The equation $ x + y = 60 $ can be re-written as:
 $
  x + y = 60 \\
  x = 60 - y \\
  $
Let the given function $ x{y^3} $ be F such that $ F = x{y^3} $ .
Substitute the value of ‘x’ in the function F, we get
\[
\Rightarrow F = x{y^3} \\
   = (60 - y){y^3} \\
 \]
Differentiate the function F with respect to ‘x’, we get
 $
\Rightarrow \dfrac{{dF}}{{dy}} = \dfrac{d}{{dy}}\left( {(60 - y){y^3}} \right) \\
   = \dfrac{d}{{dy}}\left( {60{y^3} - {y^4}} \right) \\
   = 180{y^2} - 4{y^3} - - - - (ii) \\
   = {y^2}(180 - 4y) \\
  $
Now, to determine the value of ‘x’ and ‘y’ such that $ x{y^3} $ is maximum, equate the differentiation of the function to zero and evaluate the value of ‘y’.
 $
\Rightarrow {y^2}(180 - 4y) = 0 \\
\Rightarrow {y^2} = 0{\text{ and }}180 - 4y = 0 \\
\Rightarrow y = 0{\text{ and }}y = \dfrac{{180}}{4} = 45 \\
  $
Hence, we get two values of ‘y’, but we need the maximum value so, neglecting y=0.
Now, substituting the value of y as 45 in the equation $ x = 60 - y $ to determine the value of x.
 $
  x = 60 - y \\
   = 60 - 45 \\
   = 15 \\
  $
Hence, the values of ‘x’ and ‘y’ such that $ x + y = 60 $ and $ x{y^3} $ is maximum is 15 and 45 respectively.

Note: To check whether the evaluated value is correct or not, again differentiate the equation (ii) with respect to ‘y’.
 $
\Rightarrow \dfrac{{{d^2}F}}{{d{y^2}}} = \dfrac{d}{{dy}}\left( {180{y^2} - 4{y^3}} \right) \\
   = 360y - 12{y^2} \\
  $
Substitute the value of y as 45 in the above equation
 $
\Rightarrow \dfrac{{{d^2}F}}{{d{y^2}}} = 360 \times 45 - 12{\left( {45} \right)^2} \\
   = 16200 - 24300 \\
   = - 8100 \\
  $
Here, the negative sign indicates that the value of ‘y’ calculated is correct and will give the maximum value of $ x{y^3} $ .
However, if the value of $ \dfrac{{{d^2}F}}{{d{y^2}}} $ comes negative then, it means that the corresponding value of ‘y’ results in minimum value.