Find the minimum of the function $y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$

Question

Find the minimum of the function $$y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$$

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By computer I found $\min_y=3;$ then I will prove $y\ge 3$. After squaring we got $$(x+2)(178-37x)\ge 0\quad \forall -2\le x\le 5.$$

My idea is not beautiful so I need some other solution. Thanks!

Answer 1

Let $f(x)=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}.$

Thus, $f$ is a concave function, which gives $$\min_{[-2,5]}f=\min\{f(-2),f(5)\}=f(-2)=3.$$ I used the following obvious statement.

Let $g(x)=\sqrt{f(x)},$ where $f$ is a concave twice differentiable function on $\{x|f(x)\geq0\}.$

Prove that $g$ is a concave function.

Answer 2

Let \begin{align*} f_1(x)&=\sqrt{-x^2+4x+21}\\ f_2(x)&=\sqrt{-x^2+3x+10}. \end{align*} An examination of the domains of each of these functions separately shows that \begin{align*} \mathcal{D}(f_1)&=[-3,7]\\ \mathcal{D}(f_2)&=[-2,5]. \end{align*} If we define $f(x)=f_1(x)+f_2(x),$ then the above forces the domain $\mathcal{D}(f)=[-2,5].$ Therefore, we must evaluate the function $f$ at $-2$ and $5$. However, there may be critical points inside the domain. Differentiating $f$ yields $$f'(x)=\frac{2-x}{\sqrt{-x^2+4x+21}}+\frac{3/2-x}{\sqrt{-x^2+3x+10}}. $$ Setting this equal to zero yields the following: \begin{align*} (2-x)\sqrt{-x^2+3x+10}&=(x-3/2)\sqrt{-x^2+4x+21}\\ (2-x)^2(-x^2+3x+10)&=(x-3/2)^2(-x^2+4x+21)\\ &\vdots\\ 51x^2-104x+29&=0\\ x&=\frac{29}{17},\;\frac{1}{3}, \end{align*} both of which are in $\mathcal{D}(f).$ However, plugging $1/3$ into $f'(x)$ shows that we picked up a spurious root by squaring. Evaluating, then, we have \begin{align*} f(-2)&=3 \\ f\left(\frac{29}{17}\right)&=6\sqrt{2}\approx 8.49 \\ f(5)&=4. \end{align*} This proves that the minimum occurs at $x=-2,$ and has a value of $y=3.$

Answer 3

Here is a simple demonstration -- not using calculus or concavity arguments --

that the minimum of $y=\sqrt{-x^2+4x+21}+\sqrt{-x^2+3x+10}$ is $3$.

Note that $y=\sqrt{-(x-7)(x+3)}+\sqrt{-(x-5)(x+2)}=\sqrt{5^2-(x-2)^2}+\sqrt{\left(\dfrac72\right)^2-\left(x-\dfrac32\right)^2}.$

So $y$ is defined when $-2\le x\le5$ or $\left(x-\dfrac32\right)^2\le\left(\dfrac72\right)^2. $

When that is the case, $-2\le x\le 6$ or $\left(x- 2\right)^2\le4^2. $

Therefore, $y= \sqrt{5^2-(x-2)^2}+\sqrt{\left(\dfrac72\right)^2-\left(x-\dfrac32\right)^2}\ge\sqrt{5^2-(x-2)^2}\ge\sqrt{5^2-4^2}=3,$

and note that $y=3$ when $x=-2$.