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A rectangle is to be placed in the first quadrant, with one side on the x-axis and one side on the y-axis so that the rectangle lies below the parabola y=-3x^2+4. What is the maximum area of this rectangle?

 

a) 16/9

b) 3 

c) 2/3 

d) 2 

 Mar 30, 2018
 #1
avatar+100595 
+2

Mmmm...let's see if I know this one.....

 

Let the width of the rectangle  = x

And let the height  =  -3x^2 + 4

 

So....the area, A  =  width * height   =  x (-3x^2 + 4) = -3x^3 + 4x

 

Take the derivative of A    and set to  0

 

 

A'   =  -9x^2  +  4

 

-9x^2 +  4  =  0

-9x^2  =   -4          divide both sides by -9

x^2  =  4/9             take the positive root

x = 2/3     =  width

 

And  the height is given  by:

 -3(2/3)^2  + 4   = 

-3(4/9) + 4

-4/3 + 4

8/3

 

So....the max area  is

 

(2/3)(8/3)  =   16/9   units^2

 

 

cool cool cool

 Mar 30, 2018

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