This might be too easy

Algebra Level 4

Let $a\geq b>c>0 ; a+b = 16$ , find the maximum value of $\sqrt{c(b-c)}+\sqrt{c(a-c)}.$

The answer is 8.

1 solution

Rishabh Jain
Jan 28, 2016

Applying $\color{forestgreen}{\text{Cauchy Scwarz inequality}}$ $\large A=\sqrt{c(b-c)}+\sqrt{c(a-c)}\\ \leq \sqrt {(c(b-c)+c(a-c))(1+1)}$ $=\sqrt{2(-2c^2+c(a+b))}$ $=\sqrt{-4(c-4)^2+64}~~~\color{#D61F06}{\small{(Using~a+b=16)}}$ which has a maximum value of $\Large 8$ when c=4, also a=b=8.

The coefficient of $c^2$ in the third line, should be $-2$ and NOT just $-1$ . In the 4th line, the $-2$ should be $-4$ .

Bob Kadylo - 5 years, 1 month ago

A typo....Thanks... Corrected....

Rishabh Jain - 5 years, 1 month ago

This might be too easy

The answer is 8.

1 solution

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