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

Savannah Anderson · Accepted Answer

#49/12" sq.unit."#

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.!

Avery Alexander · Answer

undefined To find the largest possible area for the rectangle formed by perpendiculars drawn from point P to both the x-axis and y-axis, we need to maximize the area.

Let's denote the coordinates of point P as $ (x, y) $. Since point P lies on the line $ y = 7 - 3x $, we can substitute $ y $ with $ 7 - 3x $ in the equation of the area of the rectangle to express it solely in terms of $ x $.

The area $ A $ of the rectangle is given by the product of the lengths of its sides:

$$ A = x \times y = x \times (7 - 3x) = 7x - 3x^2 $$

To maximize the area, we need to find the critical points of the function $ A(x) = 7x - 3x^2 $ by taking its derivative and setting it equal to zero:

$$ A'(x) = 7 - 6x $$
$$ 7 - 6x = 0 $$
$$ x = \frac{7}{6} $$

To determine whether this critical point corresponds to a maximum or minimum, we can use the second derivative test. The second derivative of $ A(x) $ is negative, indicating that the critical point $ x = \frac{7}{6} $ corresponds to a maximum.

Therefore, the largest possible area for the rectangle is obtained when $ x = \frac{7}{6} $. We can then find the corresponding value of $ y $ by substituting $ x = \frac{7}{6} $ into the equation $ y = 7 - 3x $:

$$ y = 7 - 3 \times \frac{7}{6} = 7 - \frac{7}{2} = \frac{7}{2} $$

Thus, the largest possible area for the rectangle is achieved when $ x = \frac{7}{6} $ and $ y = \frac{7}{2} $. Finally, we calculate the area of the rectangle:

$$ A = \frac{7}{6} \times \frac{7}{2} = \frac{49}{12} $$

So, the largest possible area for the rectangle is $ \frac{49}{12} $ square units.