3
$\begingroup$

My question is: Prove-

If $a,b$ are two positive real numbers such that their sum is $a+b=k$. Then the product $ab$ is maximum if and only if $a=b=\displaystyle\frac{k}{2}$.

I proved the above theorem above using $(a+b)^2$ but I want to know how can it be proved using A.M. G.M. inequality.

$\endgroup$
9
$\begingroup$

AM-GM inequality: $\displaystyle\frac{a + b}{2} \geq \sqrt{ab}$, with equality if and only if $a = b$.

Substituting $a + b = k$ and taking squares gives us $$\frac{k^2}{4} \geq ab.$$ So the product $ab$ is bounded from above by a constant, and this upper bound can be achieved only by taking $a = b = \frac{k}{2}$.

$\endgroup$
2
$\begingroup$

This is a maximization problem:

$a+b=k$. Maximize $ab$.

$a=k-b$, so $ab=(k-b)b$.

$f(b)=(k-b)b$ is maximal if $$\frac{df}{db}=0$$

$\frac{df}{db}=(k-b)-b=k-2b$

This equals $0$ if and only if $k=2b$. Therefore, $$a=b=\frac{k}{2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.