Question

Divide the number 84 into two parts such that the product of one part and square of other is maximum.

Answer 1

Hint: Assume the two numbers in such a way that their sum is 84. Then, form an expression from the given condition and apply the rule of derivatives to find the maximum value of the expression and hence the required numbers.

Complete step-by-step answer:
Let the required numbers be x and 84-x.
Let $P=x(84-x)^2$
We have to find the numbers such that the value of P is maximum.
Differentiating P w.r.t. x, we get,
$\dfrac{dP}{dx}=\dfrac{d}{dx}[x(84-x)^2]$
Applying product rule of differentiation, we get,
$\dfrac{dP}{dx}=(84-x)^2+2x(84-x)(-1)$
Simplifying, we get,
$\dfrac{dP}{dx}=(84-x)^2-2x(84-x)$
Again, differentiating it w.r.t. x, we get,
$\dfrac{d^2P}{dx^2}=\dfrac{d}{dx}[(84-x)^2-2x(84-x)]$
Simplifying, we get,
$\dfrac{d^2P}{dx^2}=\dfrac{d}{dx}[(84-x)^2]-\dfrac{d}{dx}[2x(84-x)]$
$\implies \dfrac{d^2P}{dx^2}=2(84-x)(-1)-[2x(-1)+(84-x)(2)]$
$\implies \dfrac{d^2P}{dx^2}=-2(84-x)-[-2x+2(84-x)]$
$\implies \dfrac{d^2P}{dx^2}=-2(84-x)+2x-2(84-x)$
$\implies \dfrac{d^2P}{dx^2}=-4(84-x)+2x$
Now, to find maxima or minima, we put $\dfrac{dP}{dx}=0$.
Thus, $\dfrac{dP}{dx}=(84-x)^2-2x(84-x)=0$
$\implies (84-x)^2-2x(84-x)=0$
Rearranging the terms, we get,
$(84-x)^2=2x(84-x)$
$\implies (84-x)=2x$
Again, rearranging the terms, we get,
$2x+x=84$
$\implies 3x=84$
$\implies x =28$
Now, we find the value of $\dfrac{d^2P}{dx^2}$ at x =28. If $\dfrac{d^2P}{dx^2}$> 0 at x = 28, then x = 28 is a point of minima and if $\dfrac{d^2P}{dx^2}$ < 0 at x = 28, then x = 28 is a point of maxima.
Now, $\dfrac{d^2P}{dx^2}=-4(84-x)+2x$
At x = 28,
$\dfrac{d^2P}{dx^2}=-4(84-28)+2(28)=-224+56=-168<0$
As $\dfrac{d^2P}{dx^2}$< 0, x = 28 is a point of maxima.
Now, if x =28 then, 84-x = 56.
Hence, the required numbers are 28 and 56.
Note: This type of question is an application of derivatives. In this type of question where we have to find the maximum or minimum value of a function, we use a second derivative test. Firstly, we find the first derivative of the function and by equating it to zero, we get the critical points. Putting the value of the critical point in the second derivative, if the resulting value is less than zero, the critical point is a point of maximum and vice-versa.