Another Calc 1 Question?

Question

A rectangle has its base on the x axis and its upper two vertices on the parabole y=12-x^2. What is the largest area the rectangle can have and what are its dimensions.
Please describe how you arrive at the answer

Answer 1

The parabola y=12-x^2, if you graph it, has a vertex at (0,12) and opens downwards, crossing the x axis at sqrt(12) and -sqrt(12). So the center of the base of whatever rectangle you draw is going to have its center at (0,0). At any point x, the value of y is going to be 12-(x^2). So for a given value of x, the length of the base is going to be 2x (as it stretches for "x" on either side of the y axis) and the height is 12-(x^2). This means the area of a rectangle described like this will be 2x(12-(x^2)).

Multiply this out, and you get
2x(12-(x^2)) =
24x - 2x(x^2) =
24x - 2x^3

We now have to find the value of x where this is going to be at a max. We want to find out when the rate of change in volume hits a peak, so we take the derivative and set it equal to zero:

24 - 6x^2 = 0
24 = 6x^2
4 = x^2

So x = 2 or -2. It doesn't matter which one we pick, because a rectangle with a corner at x=2 has to have one at x=-2 also. This describes a rectangle whose base is 4 units long. To get the height, just plug 2 into the parabola equation. You get 12 - (2)^2 = 12 - 4 = 8. So the height is 8, and the area is 8*4 = 32.

Answer 2

The vertices on the x axis must be symetrical to the origin.So if you call z the length of the base,the coordinates are z/2;0 and
-z/2,0
Putting z/2 in the equation of the parabola you get the height of the rectangle
y=12-z^2/4 and the area would be

A = z * ( 12-z^2/4)
dA/dz = 1/2 [12-z^2/4 - z(z/2) = 1/2 (48 -3z^2)/4
dA/dz = 0 ==> 48-3z^2 =0 z^2 =16 and z = 4 (take the positive
root).The sign of the derivative change from + to- so it is a maximum
The sides are 4 and 8

Answer 3

Okay, the area is 2xy, which is what you want to maximize.
So, that gives you , using substitution, 2x((12-x^2)) = 24x-2x^3.
So, differinate and set to 0, giving you 24-6x^2=0, or 24=6x^2.
So, 4=x^2, so, x=2.

So, the dimensions are 2*2 and (12-2^2) = 4 by 8 , or 4*8=32, which is area.