Question

Find two numbers whose sum is 24 and whose product is as large as possible.

Answer 1

Let first number = x then second number = 24 - x.

According to the question, their product p = x(24 - x) = 24x - x²               ..…(1)

For highest and lowest value, ${\displaystyle \frac{dp}{dx}=0}$

On differentiating equation (1) with respect to x,

${\displaystyle \frac{dp}{dx}=24-2x}$

${\displaystyle \Rightarrow 0=24-2x}$

${\displaystyle \Rightarrow 2x=24}$

${\displaystyle \therefore x=12}$

Again differentiating equation (1) with respect to x,

${\displaystyle \frac{d^{2}p}{{dx}^2}=-2}$     (negative value)

${\displaystyle {\left(\frac{d^{2}p}{{dx}^2}\right)}_{x=12}=-2<0}$

p has a masimum value at x = 12

So, the requied number are 12 and 24 - 12 = 12