Question

Two vertices of a rectangle are on the positive x-axis. The other two vertices lie on the lines $y=4x$ and $y=-5x+6$. Then the maximum area of the rectangle is?

• A. $\frac{2}{3}$
• B. $\frac{2}{4}$
• C. $\frac{1}{3}$
• D. $\frac{4}{3}$

Answer 1

Answer: (A)

$\frac{2}{3}$

Let the two vertex lying be $(a,0)$ and $(b,0)$ where $a<b$

Since $y=4x$ is containing coordinate corresponding to $(a,0)$ ; we get

Point to be $(a,4a)$

and the other with $y=-5x+6$ , we get point as $(b,-5b+6)$

Now as $b-a=-5b+6-4a$ (Length)

$a+2d=2$

Area of Rectangle$=$ Length $\times$ breadth $=(b-a)\left(4a\right)$

Area$=(b-a)\left(4a\right)=(\frac{2-a}{2}-a)\left(4a\right)=(2-3a)\left(2a\right)$

Area $=4a-6a^2$

For Maxima/Minima,

$\frac{d\left(Area\right)}{da}=0\Rightarrow 4-12a=0a=\frac{1}{3}$

Area$=(4a-{6a}^2)=\frac{4}{3}-\frac{6}{9}=\frac{2}{3}$ sq. unit