Minimum value of $4r + 5 r^2$

158 Views Asked by At

We can say by quadratic formula that min value will occur at $r=-2/5$.

However by AM-GM inequality, Min value of function will be achieved when $4r=5r^2$ (equality sign holds when each term is equal to other).

From here $r=4/5$

Where am I going conceptually wrong?

1

There are 1 best solutions below

1
On

$$4r+5r^2=5\left(r^2+\frac{4}{5}r\right)=5\left(r+\frac{2}{5}\right)^2-\frac{4}{5}\geq-\frac{4}{5}.$$ The equality occurs for $r=-\frac{2}{5}$, which says that $-\frac{4}{5}$ is a minimal value.