Find the maximum/minimum of function

Question

Find the minimum of the following functions $f$, (without using a calculator or calculus)

$$f = \sqrt{x + 25} + \sqrt{36 - x} + \sqrt{x}$$

I have tried the C-S-B inequality but could only find the maximum of function $f$. What am I missing here?

Answer 1

The domain is $[0,36]$.

Let $x=0$.

Hence, we get a value $11$.

We'll prove that it's a minimal value.

Indeed, we need to prove that $$\sqrt{x+25}+\sqrt{36-x}+\sqrt{x}\geq11$$ or $$\sqrt{x+25}-5-(6-\sqrt{36-x})+\sqrt{x}\geq0$$ or $$\frac{x}{\sqrt{x+25}+5}-\frac{x}{\sqrt{36-x}+6}+\sqrt{x}\geq0,$$ which is true because $$\frac{\sqrt{x}}{\sqrt{36-x}+6}\leq1,$$ which is $$6-\sqrt{x}+\sqrt{36-x}\geq0.$$ Done!

Answer 2

Recall the inequality $a + b \leq (\sqrt{a}+\sqrt{b})^2$ for $a,b\geq 0$ (since $2\sqrt{ab}\geq 0$ is missing from LHS). From this you get

$$\sqrt{a}+\sqrt{b}\geq\sqrt{a+b}$$

for $a,b\geq 0$. Use this to show that $f(x) \geq \sqrt{x+25} + 6$. Then see where this lower bound attains its minimum (Hint: the domain of $f$ is $[0,36]$ and $\sqrt{x+25} + 6$ is increasing).

Answer 3

f:=x->sqrt(x+25)+sqrt(36-x)+sqrt(x);

       f := x -> sqrt(x + 25) + sqrt(36 - x) + sqrt(x)

g:=D(f);

                      1                  1           1/2
   g := x -> 1/2 ------------ - 1/2 ------------ + -------
                 sqrt(x + 25)       sqrt(36 - x)   sqrt(x)

solve(g(x)=0,x);

           4        3          2
RootOf(5 _Z  + 56 _Z  - 2900 _Z  - 90000 _Z + 810000, index = 2)

a:=evalf(%);

                       a := 26.90515432

f(a);

                         15.40730723

This is maple output, I seriously doubt the witout calculator or anything.

Maybe for the minimum which is $f(0)=11$ there is possibility, but the maximum seems definitely not manageable manually.

Answer 4

Clearly $f$ has domain $0\leq x\leq 36$. Let $$g=\sqrt{36-x}+\sqrt{x}$$ and $$h=\sqrt{x+25}.$$ We show that $g(x)\geq 6$ for all $x\in [0, 36]$. For this we need to show that $$\sqrt{36-x}+\sqrt{x}\geq 6$$ $$i.e., \sqrt{36-x}\geq 6-\sqrt{x}$$ $$i.e., 6\sqrt{x}\geq x$$ $$i.e., x(36-x)\geq 0$$ which is true.

So $g(x)\geq 6$ for all $x\in[0, 36]$ and $g(0)=6$. Also $h(x)\geq 5$ for all $x\in [0, 36]$ and $h(0)=5$. That is both $g$ and $h$ attain minimum at $x=0$ and so $f$ has minimum at $x=0$.