Let $P(-2,3)\;\;,Q(-1,1)\;\;,R(s,t)$ and $S(2,7)$ be $4$ points in order on the parabola $y=ax^2+bx+c$. Then the coordinates of $R(s,t)$ such that the area of Quadrilateral $PQRS$ is maximum.
$\bf{My\; Try::}$ Here $P(-2,3)\;,Q(-1,1)\;\;,S(2,7)$ be the points lie on the parabola $y=ax^2+bx+c$
So Put $(x,y) = (-2,3)$ in $y=ax^2+bx+c\;,$ We get $4a-2b+c=3.................(1)$
Similarly Put $(x,y) = (-1,1)$ in $y=ax^2+bx+c\;,$ We get $a-b+c=1...........(2)$
Similarly Put $(x,y) = (2,7)$ in $y=ax^2+bx+c\;,$ We get $4a+2b+c=7...........(2)$
Now Solving These three equation , We get $a=1\;,b=1\;,c=1$
So we get equation of parabola be $y=x^2+x+1$
Now $\bf{Area\; of\; Quadrilateral\; PQRS = Area\; of \; \triangle PQR+Area\; of \; \triangle PRS}$
So Area Of Quadrilateral $$A(s,t) = \begin{vmatrix} -2& 3 & 1\\ -1& 1 & 1\\ s & t & 1 \end{vmatrix}+\begin{vmatrix} -2& 3 & 1\\ s& t & 1\\ 2 & 7 & 1 \end{vmatrix}$$
So $$A(s,t) = 2s+t+1+20+4s-4t = 6s-3t+21$$
Now $R(s,t)$ lie on $y=x^2+x+1.$ So we get $t=s^2+s+1$
So we get $$A(s) = 6s-3(s^2+s+1)+21 = -3s^2+3s+18$$
So using Derivative Test $A'(s)=-6s+3$ and $A''(s)=-6$
So for Max. or min., Put $\displaystyle -6s+3=0\Rightarrow s=\frac{1}{2}$ and $\displaystyle t=\frac{1}{4}+\frac{1}{2}+1 = \frac{7}{4}$
My Question is can we solve it any Shorter way, Means Maximise area using Inequality or Geometrically
If yes Then plz explain here.
Thanks
All that calculation was really unnecessary. Here's a fast way:
The area of the quadrilateral PQRS is equal to the sum of the areas of the triangles $\triangle RQS$ and $\triangle PQS$. And since the area of the triangle $\triangle PQS$ is fixed, we need only to maximise the area of the $\triangle RQS$
Additionally since the base QS of the $\triangle RQS$ is also fixed, we need to just maximise the height corresponding to it. To do so we find the point R on the parabola where the tangent is parallel to the base QS.
Doing so, we have $y'=2x+1=2\implies x=\frac 12$ and therefore $R\equiv(\frac 12,\frac 74)$