Question

If sum of two numbers is $3$ ,then maximum value of the product of first and the square of second is
$\left( 1 \right){\text{ }}4$
$\left( 2 \right){\text{ 3}}$
$\left( 3 \right){\text{ 2}}$
$\left( 4 \right){\text{ 1}}$

Answer 1

Hint : First we will form the equation from the given logic and Then find the first derivative of the equation and let it be equal to zero to find the critical values. After this, find the second derivative of the equation to check where the function is maximum and minimum with the help of critical values. As we have to find the maximum value, therefore put that critical number (at which function is maximum) in the original function to get the maximum value.

Complete step-by-step answer :
It is given to us that the sum of two numbers is three. And we have to find the maximum value of the product of the first number with the square of the second number.
Let x and y be the two numbers. Therefore, the sum of two numbers is given by $x + y = 3$ and their product will be $P = x{y^2}$ .
As the sum of numbers is given to us as $x + y = 3$ therefore by shifting y to the right hand side we get the value of our number x as $x = 3 - y$ .Therefore, now the product $P = x{y^2}$ becomes
$P = \left( {3 - y} \right){y^2}$ -------- (i)
And now we have to find the maximum value of the product $P = \left( {3 - y} \right){y^2}$ . As the value of the function at a maximum point is called the maximum value of the function .Whenever we have to find the maximum value or minimum value of the function then first differentiate the function and put it equal to zero to find the critical numbers. Then again differentiate the function and apply these critical numbers in the second derivative. By applying this if the value of the second derivative is less than zero the given function is maximum but if it is greater than zero then the function is minimum. And then to find the maximum or minimum value we need to put the value in the original function, the value at which the function is minimum or maximum. So, now let’s solve it further.
We can also write the equation (i) as $P = 3{y^2} - {y^3}$ .On differentiating it with respect to y we get
$\dfrac{{dP}}{{dy}} = \dfrac{{d\left( {3{y^2} - {y^3}} \right)}}{{dy}}$ --------- (ii)
Here we will use the derivative rule that $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ where n is any real number. Therefore, the equation (ii) becomes
$\dfrac{{dP}}{{dy}} = 6y - 3{y^2}$ ---------- (iii)
Now put $\dfrac{{dP}}{{dy}}$ equal to zero
$\Rightarrow 6y - 3{y^2} = 0$
Take $3y$ common from the equation
$\Rightarrow 3y\left( {2 - y} \right) = 0$
From here we have $y = 0$ and $y = 2$
Again differentiating the equation (iii) with respect to y we get
$\dfrac{{{d^2}P}}{{d{y^2}}} = 6 - 6y$ ------ (iv)
At $y = 0$ the equation (iv) becomes
${\left( {\dfrac{{{d^2}P}}{{d{y^2}}}} \right)_{y = 0}} = 6 - 6\left( 0 \right)$
That is ${\left( {\dfrac{{{d^2}P}}{{d{y^2}}}} \right)_{y = 0}} = 6$ which is $> 0$ .Therefore the given function is minimum at $y = 0$
Now at $y = 2$ the equation (iv) becomes
${\left( {\dfrac{{{d^2}P}}{{d{y^2}}}} \right)_{y = 2}} = 6 - 6\left( 2 \right)$
$\Rightarrow {\left( {\dfrac{{{d^2}P}}{{d{y^2}}}} \right)_{y = 2}} = 6 - 12$
That is $\Rightarrow {\left( {\dfrac{{{d^2}P}}{{d{y^2}}}} \right)_{y = 2}} = - 6$ which is $< 0$ .Therefore the given function is maximum at $y = 2$
Now put $y = 2$ in equation (i) to get the maximum value, Therefore $P = \left( {3 - y} \right){y^2}$ becomes
$P = \left( {3 - 2} \right){\left( 2 \right)^2}$
On further solving we get
$\Rightarrow P = 1 \times 4$
$\Rightarrow P = 4$
So we get the maximum value of the product as $4$ .
Hence, the correct option is $\left( 1 \right){\text{ }}4$.

Note : Keep in mind the procedure to find the maximum value and minimum value of the function and conditions where the function has maximum and minimum values. Try to understand the question that they try to ask you to find out. Remember all the derivative rules so that it will be easy for you to answer the questions.