Question

What is the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 3 − x2?

Answer 1

C. 4

Explanation:

After plotting the curves, we get the figure given below.

So, the rectangle will lie in 1st and 2nd quadrant.

Thus, let the vertex in the 1st quadrant = ( x,y ) and in 2nd quadrant = ( -x,y ).

Then, the length of the rectangle = 2x and width of the rectangle = y.

As, area of a rectangle = length × width

Therefore, area of the given rectangle, A = 2x × y

i.e.
`A=2x(3-x^(2))`

i.e.
`A=6x-2x^(3)`

Thus, differentiating with respect to x and equating to 0 gives,

`(dA)/(dx)=0`

i.e.
`6-6x^(2)=0`

i.e.
`6x^(2)=6`

i.e.
`x^(2)=1`

i.e.
`x=1,-1`

Again, differentiating with respect to x gives us
`(d^(2)A)/(dx^(2)) =-12x`.

If x = -1 ,
`(d^(2)A)/(dx^(2)) =12>0`.

If x =1 ,
`(d^(2)A)/(dx^(2)) =-12<0`. This gives us that the maximum value of the area is obtained at x = 1.

Thus, length = 2x = 2 and width = y =
`3-x^(2)` = 3- 1 = 2

So, area of the rectangle is A = 2(1) × 2 = 4.

Hence, area of the rectangle is 4
`unit^(2)`.

Answer 2

(C) 4

Explanation:

Let us first construct a rectangle ABCD, in which A is the point at lower right hand corner and B,C,D are the points marked according to A,

Now, let A=
`(p,0)`

B=
`(-p,0)`,

C=
`(-p,3-p^(2))`

and D=
`(p,3-p^(2))`

Then the area of rectangle is given as: BA×AD

=
`(2p)(3-p^(2))`

A=
`6p-2p^(3)`

Taking the derivative with respect to p, we have

`A^(')`=
`6-6p^(2)`

Now,
`A^(')`=0

⇒
`6-6p^(2)`=0

⇒
`6p^(2)=6`

⇒
`p^(2)=1`

Since, wehav eto find the greater area, therefore we will take p=1.

Now, substituting the value of p in (A), we have

Greater area= A=
`6-2(1)`=
`4 sq units`