Need help with review question on test tomorow

[2Pac]

need help please. tonight if possible. I need to find the largest area of a rectangle with base on the x-axis and upper corners touch the parabola with the equation y=12-x^2
thanks

 

[AKG]

Well, you know the rectangle will be symmetrical about the y-axis. That means, if it touches the parabola at x = -a, it will touch at x = a as well. The rectangle will have an area of $A = bh = (2x)(12 - x^2)$. Hopefully you see why this is. Now, it's simply a matter of finding the x value which gives a maximum for A, and I'm sure you know how to do this.

 

[M98Ranger]

Sure! I would be happy to help with your review question. To find the largest area of a rectangle with a base on the x-axis and upper corners touching the parabola y=12-x^2, we can use the formula for the area of a rectangle, which is A = length x width.

In this case, the length of the rectangle will be the x-coordinate of the upper corners, and the width will be the distance between the two upper corners. Since the upper corners touch the parabola, we can set the x-coordinate of the upper corners equal to the x-coordinate of the vertex of the parabola, which is x=0.

To find the distance between the two upper corners, we can use the distance formula, which is d = √((x2-x1)^2 + (y2-y1)^2). In this case, we can set one of the upper corners to be (0,12) and the other upper corner to be (x,12-x^2). Plugging these values into the distance formula, we get d = √((x-0)^2 + ((12-x^2)-12)^2) = √(x^2 + x^4).

Now, we can plug in these values into the area formula A = length x width. A = (0)(√(x^2 + x^4)) = 0.

Since the area cannot be negative, we know that the largest area of the rectangle is when x=0, which means the rectangle is actually a square with sides of length 0. Therefore, the largest area of the rectangle is 0 units squared.

I hope this helps with your review and good luck on your test tomorrow! Let me know if you have any other questions.