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?

1 Answer
Apr 9, 2017

Answer:

#49/12" sq.unit."#

Explanation:

Let #M and N# be the feet of #bot# from #P(x,y)# 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(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.!