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Question: Find two positive integers x and y such that \( x + y = 60 \) and \( x{y^3} \) is maximum....

Find two positive integers x and y such that x+y=60x + y = 60 and xy3x{y^3} is maximum.

Explanation

Solution

Hint : In this question, we need to evaluate the value of ‘x’ and ‘y’ such that the sum of the numbers be 60 and xy3x{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=60x + y = 60 can be re-written as:
x+y=60 x=60−y  x + y = 60 \\\ x = 60 - y \\\
Let the given function xy3x{y^3} be F such that F=xy3F = x{y^3} .
Substitute the value of ‘x’ in the function F, we get

⇒F=xy3 =(60−y)y3 \Rightarrow F = x{y^3} \\\ = (60 - y){y^3} \\\

Differentiate the function F with respect to ‘x’, we get
⇒dFdy=ddy((60−y)y3) =ddy(60y3−y4) =180y2−4y3−−−−(ii) =y2(180−4y)  \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 xy3x{y^3} is maximum, equate the differentiation of the function to zero and evaluate the value of ‘y’.
⇒y2(180−4y)=0 ⇒y2=0 and 180−4y=0 ⇒y=0 and y=1804=45  \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−yx = 60 - y to determine the value of x.
x=60−y =60−45 =15  x = 60 - y \\\ = 60 - 45 \\\ = 15 \\\
Hence, the values of ‘x’ and ‘y’ such that x+y=60x + y = 60 and xy3x{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’.
⇒d2Fdy2=ddy(180y2−4y3) =360y−12y2  \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
⇒d2Fdy2=360×45−12(45)2 =16200−24300 =−8100  \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 xy3x{y^3} .
However, if the value of d2Fdy2\dfrac{{{d^2}F}}{{d{y^2}}} comes negative then, it means that the corresponding value of ‘y’ results in minimum value.