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

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06-01-2019
Mathematics

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

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boffeemadrid boffeemadrid · Answer 1 · 2019-01-16T05:40:15+01:00

Answer:

(C) 4

Step-by-step 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= [tex](p,0)[/tex]

B=[tex](-p,0)[/tex],

C=[tex](-p,3-p^{2})[/tex]

and D=[tex](p,3-p^{2})[/tex]

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

=[tex](2p)(3-p^{2})[/tex]

A=[tex]6p-2p^{3}[/tex]

Taking the derivative with respect to p, we have

[tex]A^{'}[/tex]=[tex]6-6p^{2}[/tex]

Now, [tex]A^{'}[/tex]=0

⇒[tex]6-6p^{2}[/tex]=0

⇒[tex]6p^{2}=6[/tex]

⇒[tex]p^{2}=1[/tex]

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=[tex]6-2(1)[/tex]=[tex]4 sq units[/tex]

wagonbelleville wagonbelleville · Answer 2 · 2019-01-16T05:52:36+01:00

Answer:

C. 4

Step-by-step 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. [tex]A=2x(3-x^{2})[/tex]

i.e. [tex]A=6x-2x^{3}[/tex]

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

[tex]\frac{dA}{dx}=0[/tex]

i.e. [tex]6-6x^{2}=0[/tex]

i.e. [tex]6x^{2}=6[/tex]

i.e. [tex]x^{2}=1[/tex]

i.e. [tex]x=1,-1[/tex]

Again, differentiating with respect to x gives us [tex]\frac{d^{2}A}{dx^{2}} =-12x[/tex].

If x = -1 , [tex]\frac{d^{2}A}{dx^{2}} =12>0[/tex].

If x =1 , [tex]\frac{d^{2}A}{dx^{2}} =-12<0[/tex]. This gives us that the maximum value of the area is obtained at x = 1.

Thus, length = 2x = 2 and width = y = [tex]3-x^{2}[/tex] = 3- 1 = 2

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

Hence, area of the rectangle is 4 [tex]unit^{2}[/tex].