Find the global minimum of $ab+a+b$ subject to $a^2+b^2=25$

Question

Let $a,b$ be real numbers. İf $a^2+b^2=25$ find minimum value of $ab+a+b$

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My solution: $b=\sqrt{25-a^2}$
$f(a)=a\sqrt{25-a^2}+a+\sqrt{25-a^2}$ and I use $f'(a)=0$ But this way long and difficult to me. Is there any easy solution? Thanks.

Answer 1

Hint 1: Notice that $$(a+b+1)^2=a^2+b^2+1+2ab+2a+2b=26+2(ab+a+b)$$ so there is an equivalent much simpler function to minimize.

Hint 2: Pythagorean triplets.

Answer 2

Another way to do these kind of problems is using lagrange multipliers

That is, write $\mathcal{L}(a,b,\lambda)=ab+a+b+\lambda(a^2+b^2-25)$ and solve the system of equations $$\begin{cases}\frac{\mathrm{d}}{\mathrm{d}a}\mathcal{L}=0\\\frac{\mathrm{d}}{\mathrm{d}b}\mathcal{L}=0\\\frac{\mathrm{d}}{\mathrm{d}\lambda}\mathcal{L}=0\end{cases}$$

In our case, this would be the system $$\begin{cases}b+1+2\lambda a=0\\a+1+2\lambda b=0\\a^2+b^2-25=0\end{cases}$$

which is easier to solve.

Answer 3

You might consider using substitutions as another way.

Let $\thinspace a=m+n, \thinspace b=m-n$ , then you have :

$$ \begin{align}&a^2+b^2=25\\ \iff &m^2+n^2=\frac {25}{2}\end{align} $$

This leads to :

$$ \begin{align}ab+a+b&=m^2-\left(\frac {25}{2}-m^2\right)+2m\\ &=2m^2+2m-\frac {25}{2}\\ &=\frac 12\left(2m+1\right)^2-13\\ &≥-13\thinspace .\end{align} $$

Answer 4

Let $\thinspace a+b=m, \thinspace ab=n\thinspace $ then you have :

$$m^2-2n=25$$

This implies that :

$$ \begin{align}m+n&=m+\frac {m^2-25}{2}\\ &=\frac 12\left(m+1\right)^2-13\\ &≥-13\thinspace .\end{align} $$

Answer 5

We have that

$$a^2+b^2=25 \iff (a+b)^2=25+2ab \iff ab=\frac12(a+b)^2-\frac{25}2$$

and

$$ab+(a+b)=\frac12(a+b)^2+(a+b)-\frac{25}2=\frac12((a+b)+1)^2-13$$