p228 q20c
given area of triangle $\displaystyle A = \frac{2(r^2+r+1)}{(1+r)^2}$
find the min. area
i put r --> infinty then A = 2 but it is wrong. the answer of my book says A=3/2
thanks in advance
$\displaystyle A = 2\left(1 - \frac{r}{(1+r)^2}\right)$
A achieves minimum when $\displaystyle \frac{r}{(1+r)^2}$ is maximum.
But $\displaystyle (1+r)^2 \geq 4r \Rightarrow \frac{r}{(1+r)^2} \leq \frac14 \Rightarrow \left(1 - \frac{r}{(1+r)^2}\right) \geq \frac34 \Rightarrow A \geq \frac32$
With equality at r=1.
So minimum of A is achieved at r=1.And that is A = 3/2.
To prove that $\displaystyle 1 + 2r + r^2 \geq 4r$ you only have to prove that $\displaystyle 1 + r^2 \geq 2r$.
Draw a graph of $\displaystyle y = x^2 + 1$ and $\displaystyle y = 2x$ and it's not hard to see (and then prove) that the line $\displaystyle y = 2x$ is tangent to $\displaystyle y = x^2 + 1$ at (1, 2) ........