If two positive integers x and y such that $x + 2y = 60$ then find the max vaue of xy?

Question

Can anyone please solve the question with steps

I did this by just assuming that let give both $x$ and $2y$ equal value $30$, $30$

so $x\times y = 30\times15 = 450$ (By the way $450$ is the correct answer)

Can anyone do this problem in steps.

Answer 1

Hint:$$xy=(60-2y)y=60y-2y^2=-2(y-15)^2+2\times15^2.$$

Answer 2

No matter what $x$ and $2y$ are, as long as they are positive, we always have $$ (\sqrt x-\sqrt{2y})^2\geq0 $$ with equality if and only if $\sqrt x=\sqrt{2y}$. Expanding the square, and rearranging, we get $$ x+2y\geq2\sqrt{2xy} $$ From here it is not difficult to work out the maximal possible value of $xy$. Also note that none of these operations change when the inequality is an equality.

Answer 3

Let $x+2y=60$ and $x=30-n$, $2y=30+n$

$xy=\frac{(30-n)(30+n)}{2}=\frac{30^2-n^2}{2}$

for max{xy}, $n=0$

$xy=450$