Find the max value of $\sqrt{5x - x^2} + \sqrt{18 + 3x - x^2}$

Question

I have this expression:

$C = \sqrt{5x - x^2} + \sqrt{18 + 3x - x^2}$

And I need to find the max value of $C$, can anyone help me?

I tried something like this:

$$ C^2 = 18 + 8x - 2x^2 + 2\sqrt{5x - x^2}\sqrt{18 + 3x - x^2} \Rightarrow C^2 \leq 18 + 8x - 2x^2 + 18 + 8x - 2x^2 = 36 + 16x - 4x^2 $$

But if I use this way equality can't occur.

Answer 1

Since $\;5x-x^2\geqslant0\;$ for all $x\in\left[0,5\right]\;$ and $\;18+3x-x^2\geqslant0\;$ for all $x\in\left[-3,6\right]\,,\;$ then the domain of the expression

$E(x)=\sqrt{5x-x^2}+\sqrt{18+3x-x^2}$

is the interval $\;\left[0,5\right].$

I will find all real positive values of $\;C\;$ such that

$\sqrt{5x-x^2}+\sqrt{18+3x-x^2}\leqslant C\;\quad\forall\,x\in\left[0,5\right].\quad\color{blue}{(*)}$

First of all,$\,$ since $\,E(1)=2+\sqrt{20}>6\;,\;$ it follows that $\;C>6\;.$

Moreover, for any $\,x\in\left[0,5\right]\;$ it results that

$\sqrt{18+3x-x^2}=\sqrt{8+\left(5-x\right)\left(x+2\right)}\geqslant\sqrt{8}>0\;,$

$C-\sqrt{5x-x^2}=C-\sqrt{\dfrac{25}4-\left(x-\dfrac52\right)^{\!2}}>6-\dfrac52>0\,.$

The inequality $\;(*)\;$ is equivalent to

$\underset{\overbrace{\text{the LHS is positive}}}{\sqrt{18+3x-x^2}}\leqslant\underset{\overbrace{\text{the RHS is positive}}}{C-\sqrt{5x-x^2}}\;\quad\forall\,x\in\left[0,5\right]\;,$

$18\!+\!3x\!-\!x^2\leqslant C^2\!-\!2C\sqrt{5x\!-\!x^2}\!+\!5x\!-\!x^2\quad\forall\,x\in\left[0,5\right]\,,$

$18-2x-C^2\leqslant -2C\sqrt{5x-x^2}\;\quad\forall\,x\in\left[0,5\right]\,,$

$\underset{\overbrace{\text{the LHS is positive because }C>6}}{2x+C^2-18}\geqslant\underset{\overbrace{\text{the RHS is nonnegative}}}{2C\sqrt{5x-x^2}}\;\quad\forall\,x\in\left[0,5\right]\,,$

$4x^2+C^4+324+4C^2x-72x-36C^2\geqslant20C^2x-4C^2x^2\,,$

$4\!\left(C^2\!+1\right)\!x^2-8\!\left(2C^2\!+9\right)\!x+\left(C^2\!-18\right)^2\geqslant0\;,$

$4\!\left(C^2\!+1\right)\!\left[x^2-2\!\left(\dfrac{2C^2\!+9}{C^2\!+1}\right)\!x+\left(\dfrac{2C^2\!+9}{C^2\!+1}\right)^2\right]+\left(C^2\!-18\right)^2-\dfrac{4\left(2C^2\!+9\right)^2}{C^2\!+1}\geqslant0\qquad\forall\,x\in\left[0,5\right]\,,$

$4\!\left(C^2\!+\!1\right)\!\left(\!x-\dfrac{2C^2\!+\!9}{C^2\!+\!1}\right)^{\!2}\!+\left(C^2\!-\!18\right)^2-\dfrac{4\left(2C^2\!+\!9\right)^2}{C^2\!+\!1}\geqslant0$

for all $\,x\in\left[0,5\right]\;.\quad\color{blue}{(**)}$

So far I have proved that the inequalities $\;(*)\;$ and $\;(**)\;$ are equivalent.

Since $\,C>6\;,\;$ it results that

$2<\dfrac{2C^2+9}{C^2+1}=2+\dfrac7{C^2+1}<2+\dfrac7{37}<\dfrac52\;.$

From the inequality $\,(**)\,,$ for $\;x=\dfrac{2C^2+9}{C^2+1}\in\left]2,\dfrac52\right[\;,\;$ it follows that

$\left(C^2-18\right)^2-\dfrac{4\left(2C^2+9\right)^2}{C^2+1}\geqslant0\;,$

which is equivalent to

$\left(C^2+1\right)\left(C^2-18\right)^2-4\left(2C^2+9\right)^2\geqslant0\;,$

$\left(C^2+1\right)\left(C^4-36C^2+324\right)-4\left(4C^4+36C^2+81\right)\geqslant0\;,$

$C^6-51C^4+144C^2\geqslant0\;,$

$C^2\left(C^4-51C^2+144\right)\geqslant0\;,$

$C^2\left(C^2-3\right)\left(C^2-48\right)\geqslant0\;.$

Since $\;C>6\;,\;$ the last inequality is equivalent to

$C^2-48\geqslant0\;,$

$C\geqslant4\sqrt3\;.$

Therefore, all real positive values of $\;C\;$ such that

$E(x)=\sqrt{5x-x^2}+\sqrt{18+3x-x^2}\leqslant C\;\quad\forall\,x\in\left[0,5\right]\;,$

are $\;\,C\geqslant4\sqrt3\;.$

Hence the set of upper bounds of the expression $\;E(x)\;$ is $\;\big[4\sqrt3,+\infty\big[\;$ and the supremum value of $\;E(x)\;$ is its least upper bound that is $\;C^*=4\sqrt3\;.$

Since $\;\left(C^{*2}\!-\!18\right)^2-\dfrac{4\left(2C^{*2}\!+\!9\right)^2}{C^{*2}\!+\!1}=0\;,\;$ the inequality

$E(x)=\sqrt{5x-x^2}+\sqrt{18+3x-x^2}\leqslant C^*\;\quad\forall\,x\in\left[0,5\right]\;$

is equivalent to

$4\!\left(C^{*2}\!+\!1\right)\!\left(\!x-\dfrac{2C^{*2}\!+\!9}{C^{*2}\!+\!1}\right)^{\!2}\geqslant0\;\quad\forall\,x\in\left[0,5\right]\;.$

So there exists $\;x^*\!=\!\dfrac{2C^{*2}\!+\!9}{C^{*2}\!+\!1}=\dfrac{15}7\in\big[0,5\big]\;$ such that

$E(x^*)=C^*=4\sqrt3\;.$

Consequently $\;C^*=4\sqrt3\;$ is not only the supremum value of the expression $\;E(x)\,,\,$ but also its maximum value.

Answer 2

Let us imagine an $xy$-plane. Draw a half circle over the $x$-axis, with the center $(5/2,0)$ and radius $5/2$. Draw a half circle under the $x$-axis, with the center $(3/2,0)$ and radius $9/2$. Imagine a vertical line $\ell_x$ going through a point $(x, 0)$. The value of your function at $x$ is, the length of the segment $s_x$ cut off from $\ell_x$ by these half circles. If $x$ is outside of $[0,5]$, then you see the line $\ell_x$ is not cut into a finite-length segment.

The question is, when is $s_x$ longest? It is easy to see that $x$ should be in $[3/2, 5/2]$. Imagine tangent lines (to the circles) at the endpoints of $s_x$. The segment $s_x$ becomes longest when these lines are parallel. For such $x$, two corresponding radii are also parallel. Therefore, the point $x$ should divide $[3/2, 5/2]$ into the ratio of $9/2$ to $5/2$. Hence $x=15/7$, and the maximal value is $4\sqrt{3}$.