# What is the maximum of $\ 2(a+b)-ab$ if we have: $\ a^2+b^2=8\ (a,b\ real)$

What is the maximum of $\quad 2(a+b)-ab\$ if we have: $a^2+b^2=8 \ (a,b\ real)$
My work is as follows:
According to AM-GM inequality:
$$\frac{a^2+b^2}2\ge\sqrt{a^2b^2} \Rightarrow$$
$$ab\le4$$
On the other hand , from $\ a^2+b^2=8\ ,\$we have: $\ 2(a+b)-ab=2\sqrt{8+2ab}-ab$
We can get this: $$2\sqrt{8+2ab}\le8$$
Is it possible to somehow combine these two inequalities to find the maximum value of $\ 2(a+b)-ab$ ?
I think in this solution technique we must find the minimum value of $ab\$ , but how???

• Also use $\frac{a+b}{2}\geq \sqrt{a b}$ – 1110101001 Mar 1 '16 at 5:49
• @1110101001 Plz explain more – Hamid Reza Ebrahimi Mar 1 '16 at 5:53

Another, easier method(without trigonometry and differentiation):

$$a^2+b^2 = (a + b)^2 - 2ab = 8 \rightarrow ab=\frac{(a + b)^2}{2} - 4$$

Substitute in your expression. Now you have to find the maximum of the following: $$2(a + b) - \frac{(a + b)^2}{2} + 4$$Let $t = a + b$. Since $a, b$ are arbitrary reals, then $t$ is also arbitrary real.

$$f(t) = -\frac{1}{2}t^2 + 2t + 4$$ $f(t)$ is a quadratic function, since the coefficient before $t^2$ is negative, the parabola is turned downwards. Its vertex is $t_0 = -\frac{b}{2a} = 2$, hence f(t) attains maximum value when $t = 2$ and that value is $$f(2) = 6$$

• Can my idea of using inequalities lead to the solution? – Hamid Reza Ebrahimi Mar 1 '16 at 6:36
• @HamidRezaEbrahimi From $ab = \frac{(a + b)^2}{2} - 4$ you have that $(a+b)^2 \geq 0$, hence $ab \geq -4$ – thefunkyjunky Mar 1 '16 at 6:43
• If we let $ab=-4$ in $2\sqrt{8+2ab}-ab\$ we get to the maximum value of 12!! – Hamid Reza Ebrahimi Mar 1 '16 at 6:52
• You have a mistake. The value under the square root is 0, so we get 4. – thefunkyjunky Mar 1 '16 at 6:53
• Because if you look closely at the expression, you can notice that when $ab$ increases, the value of square root increases, but then you "remove" more(because you have $-ab$). You need to find the balance between the two things. – thefunkyjunky Mar 1 '16 at 6:56

Using differentiation: Let $f(t)=2\sqrt{8+2t}-t$, then $f'(t)=\frac{2}{\sqrt{8+2t}}-1$. Then $f(t)$ increases when $-4<t<-2$ and decreases when $-2<t<4$. Thus $f(t)$ has a maximum at $t=-2$ and $f(-2)=2\sqrt{8-4}+2=6$. Since $|ab|\le 4$, $2(a+b)-ab$ has a maximum $6$.

Alternative method: Let $a=2\sqrt{2}\cos\theta$ and $b=2\sqrt{2}\sin\theta$, then \begin{align} 2(a+b)-ab&=-4(\cos\theta+\sin\theta)^2+4\sqrt{2}(\cos\theta+\sin\theta)+4\\ &=-4((\cos\theta+\sin\theta)^2-\sqrt{2}(\cos\theta+\sin\theta)-1)\\ &=-4\left(\left(\cos\theta+\sin\theta-\frac{1}{\sqrt{2}}\right)^2-\frac{3}{2}\right) \end{align} Therefore, $2(a+b)-ab$ has maximum $6$ when $\sin\theta+\cos\theta=\frac{1}{\sqrt{2}}$.

• Thank you,but the thefunky's method is simpler – Hamid Reza Ebrahimi Mar 1 '16 at 6:34

let $u=\sqrt{8+2ab} \implies ab=\dfrac{u^2-8}{2} \implies 2u-\dfrac{u^2-8}{2}=-\dfrac{(u-2)^2}{2}+6,|ab| \le \dfrac{a^2+b^2}{2}=4 \implies 0\le u \le 4$