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

Find the two positive numbers xx and yy such that x+y=60x + y = 60 and xy3x{y^3} is maximum.

Explanation

Solution

Hint : This question is based on Linear Equations as well as the Maxima and Minima. In this question one linear equation having two variables is given and also a polynomial is given whose value is maximum at some point. We have to calculate the numbers at which the value of the function satisfies the conditions stated in the question. In order to find the maximum or minimum value of the function we differentiate the function then equate it with zero and calculate the value of the variable.

Complete step-by-step answer :
Given:
The two positive numbers xx and yy satisfies this linear equation –
x+y=60x + y = 60
So, we can write this equation as –
x=60−yx = 60 - y
Let us assume the function that has a maximum value is f(x,y)f\left( {x,y} \right) .
Then, we have given,
f(x,y)=xy3f\left( {x,y} \right) = x{y^3}
Substituting the value of the xx from the first equation, we get,
f(x,y)=(60−y)y3 f(x,y)=60y3−y4 f\left( {x,y} \right) = \left( {60 - y} \right){y^3}\\\ f\left( {x,y} \right) = 60{y^3} - {y^4}
Now differentiating both sides with respect to yy we get,
\Rightarrow \dfrac{d}{{dy}}\left\\{ {f\left( {x,y} \right)} \right\\} = \dfrac{d}{{dy}}\left( {60{y^3} - {y^4}} \right)
For the maximum or minimum value of f(x,y)f\left( {x,y} \right)
\Rightarrow \dfrac{d}{{dy}}\left\\{ {f\left( {x,y} \right)} \right\\} = 0
So, by substituting we get,
⇒ddy(60y3−y4)=0 ⇒60×3y2−4y3=0 ⇒180y2−4y3=0 ⇒4y2(45−y)=0 \Rightarrow \dfrac{d}{{dy}}\left( {60{y^3} - {y^4}} \right) = 0\\\ \Rightarrow 60 \times 3{y^2} - 4{y^3} = 0\\\ \Rightarrow 180{y^2} - 4{y^3} = 0\\\ \Rightarrow 4{y^2}\left( {45 - y} \right) = 0
By solving this we get the values of yy as –
4y2=0 y=0 4{y^2} = 0\\\ y = 0
And,
(45−y)=0 y=45 \left( {45 - y} \right) = 0\\\ y = 45
Now to check whether this value of yy is either maximum or minimum, differentiating the function again with respect to yy we get,
\Rightarrow \dfrac{{{d^2}}}{{d{y^2}}}\left\\{ {f\left( {x,y} \right)} \right\\} = \dfrac{{{d^2}}}{{d{y^2}}}\left( {60{y^3} - {y^4}} \right)\\\ \Rightarrow \dfrac{{{d^2}}}{{d{y^2}}}\left\\{ {f\left( {x,y} \right)} \right\\} = \dfrac{d}{{dy}}\left( {180{y^2} - 4{y^3}} \right)\\\ \dfrac{{{d^2}}}{{d{y^2}}}\left\\{ {f\left( {x,y} \right)} \right\\} = 360y - 12{y^2}
At the point y=45y = 45 the value of this differential is –
\dfrac{{{d^2}}}{{d{y^2}}}\left\\{ {f\left( {x,y} \right)} \right\\} = 360\times(45) - 12\times {45^2} = -8100
Now since the value of the second differential at this point is negative, it means that the value of the function at this point would be maximum.
So, the function f(x,y)f\left( {x,y} \right) has a maximum value at y=45y = 45 .
Now to find the value of xx , from the first equation we have,
x+45=60 x=60−45 x=15 x + 45 = 60\\\ x = 60 - 45\\\ x = 15
Therefore, the two positive numbers at which the value of the function is maximum are 15 and 45.

Note : To check whether a function has a maxima or minima at a point we have to do a second differentiation of the function then find the value of the second differentiation at the point. If the value is negative then the function has a maxima at that point and if the value is positive then the function has a minima at that point.