# The point P lies in the first quadrant on the graph of the line y= 7-3x. From the point P, perpendiculars are drawn to both the x-axis and y-axis. What is the largest possible area for the rectangle thus formed?

Apr 9, 2017

$\frac{49}{12} \text{ sq.unit.}$

#### Explanation:

Let $M \mathmr{and} N$ be the feet of $\bot$ from $P \left(x , y\right)$ to the $X -$ Axis

and $Y -$ Axis, resp., where,

P in l={(x,y) | y=7-3x, x>0; y>0} sub RR^2....(ast)

If $O \left(0 , 0\right)$ is the Origin, the, we have, $M \left(x , 0\right) , \mathmr{and} , N \left(0 , y\right) .$

Hence, the Area A of the Rectangle $O M P N ,$ is, given by,

$A = O M \cdot P M = x y , \text{ and, using } \left(\ast\right) , A = x \left(7 - 3 x\right) .$

Thus, $A$ is a fun. of $x ,$ so let us write,

$A \left(x\right) = x \left(7 - 3 x\right) = 7 x - 3 {x}^{2.}$

For ${A}_{\max} , \left(i\right) A ' \left(x\right) = 0 , \mathmr{and} , \left(i i\right) A ' ' \left(x\right) < 0.$

$A ' \left(x\right) = 0 \Rightarrow 7 - 6 x = 0 \Rightarrow x = \frac{7}{6} , > 0.$

Also, $A ' ' \left(x\right) = - 6 , \text{ which is already } < 0.$

Accordingly, ${A}_{\max} = A \left(\frac{7}{6}\right) = \frac{7}{6} \left\{7 - 3 \left(\frac{7}{6}\right)\right\} = \frac{49}{12.}$

Therefore, the largest possible area of the rectangle is $\frac{49}{12} \text{ sq.unit.}$

Enjoy Maths.!