The base $AB$ of a Triangle is $1$ and $h$ of $C$ from $AB$ is $\leq 0.5$. Then find maximum value of product of altitudes of triangle .
Try: let in triangle $ABC$ , where $h\leq 1/2$
using area of triangle $\displaystyle \triangle=\frac{1}{2}\times 1\times h\leq \frac{1}{4}$.
Could some help me how to find product of altudes. Thanks
Assuming the altitude $h_C$ is given the problem is reduced to finding the maximum value of $h_A\cdot h_B$. As the length $AB$ is also given the area of the triangle $S=\frac{1}{2} AB\cdot h_C$ is constant. Observing that $S=\frac{1}{2} BC\cdot h_A=\frac{1}{2} AC\cdot h_B$ one obtains: $$ h_A\cdot h_B=\frac{4S^2}{AC\cdot BC}=2S\sin\hat{C}. $$ Thus the maximum value of $h_A\cdot h_B$ for given $S$ corresponds to the maximum value of $\sin\hat{C}$, which is, provided that $h_C\le \frac{1}{2}AB$, equal to $1$.
Finally one obtains: $$ \max(h_Ah_Bh_C)=\max(2S\cdot h_C)=\max(h_C^2)\cdot AB=0.25. $$