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?

Question

5418 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-09T14:14:44

$\frac{49}{12} sq.unit.$ $49/12" sq.unit."$

Let $M and N$ $M and N$ be the feet of $⊥$ $bot$ from $P (x, y)$ $P(x,y)$ to the $X -$ $X-$ Axis

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

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

If $O(0,0)$ is the Origin, the, we have, $M(x,0), and, N(0,y).$

Hence, the Area A of the Rectangle $OMPN,$ is, given by,

$A=OM*PM=xy," and, using "(ast), A=x(7-3x).$

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

$A(x)=x(7-3x)=7x-3x^2.$

For $A_(max), (i) A'(x)=0, and, (ii) A''(x)<0.$

$A'(x)=0 rArr 7-6x=0 rArr x=7/6, >0.$

Also, $A''(x)=-6," which is already "<0.$

Accordingly, $A_(max)=A(7/6)=7/6{7-3(7/6)}=49/12.$

Therefore, the largest possible area of the rectangle is $49/12" sq.unit."$

Enjoy Maths.!