Question

Find two positive numbers x and y such that x+y=60 and xy is maximum?

Answer 1

Hint: Start by finding the AM and GM of x and y which are $\dfrac{x+y}{2}$ and $\sqrt{xy}$ , respectively. Now we know that AM is always greater than or equal to GM of two positive numbers. Also, for xy to be maximum $\sqrt{xy}$ should be maximum. So, we will take AM=GM and solve to get the maximum value of xy.

Complete step-by-step answer:
Let us start the solution to the above question by finding the AM and GM of the two numbers x and y. We know that the AM of two numbers is equal to half of the sum of the two numbers while the GM is the root of the product of the two numbers.
So, AM and GM of x and y are $\dfrac{x+y}{2}$ and $\sqrt{xy}$ , respectively.
Now we know that AM is always greater than or equal to GM of two positive numbers and x and y both are positive.
$\therefore \dfrac{x+y}{2}\ge \sqrt{xy}$
Also, for xy to be maximum $\sqrt{xy}$ should be maximum. So, we will take AM=GM and solve to get the maximum value of xy.
$\therefore \dfrac{x+y}{2}=\sqrt{xy}$
Now we will put the value x+y=60. On doing so, we get
$\dfrac{60}{2}=\sqrt{xy}$
$\Rightarrow 30=\sqrt{xy}$
So, we will square both sides of the equation and we know that square of 30 is equal to 900.
$xy=900$
Therefore, the maximum value of xy is equal to 900.
Now as we have taken AP=GP, we know that the numbers for which the AP=GP are equal. So, x=y.
So, if we put x=y in x+y=60, we get
$2x=60$
$\Rightarrow x=30$
Therefore, we can conclude that x=y=30.

Note: Remember that for two positive real numbers, $AM\ge GM\ge HM$ . Also, if AM=HM=GM, we can say that both the numbers are equal and the value we are having is the maximum value of their GM and HM, as we did in the above question. If you want, you can solve the above question without using the fact the numbers are equal by solving the two equations x+y=60 and xy=900, but that would be time consuming.