How to find the width of a rectangle bounded by the x axis

[Akalan]

A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y=5−x 2 . What are the dimensions of such a rectanlge that has the greatest possible area? List them as base,height

I have no idea how to solve this. I tried graphing y=0 and the parabola, but then I dont know where to put the vertical lines. How do I approach this question?

 

[stapel]

A rectangle is inscribed with its base on the x axis and its upper corners on the parabola y=5−x 2 . What are the dimensions of such a rectanlge that has the greatest possible area? List them as base,height

I have no idea how to solve this. I tried graphing y=0 and the parabola, but then I dont know where to put the vertical lines. How do I approach this question?

Draw the axis system and the parabola. Note that the parabola is centered on the y-axis, so you only have to find the area of half of the rectangle, by working only in the first quadrant. (Once you find the maximal area in QI, you'll multiply by 2.)

Draw a rectangle that goes from the origin to some distance to the right (but still to the left of the parabola's [imath]x[/imath]-intercept), and from the [imath]x[/imath]-axis to the parabola. The rectangle's height is the vertical line down to the [imath]x[/imath]-axis where the rectangle's corner touches the parabola.

What is the width of the rectangle in the first quadrant? (Hint: The width goes from zero to [imath]x[/imath])

What is the height of the rectangle? In other words, what is the [imath]y[/imath]-value of the intersection of the rectangle and the parabola? (Hint: Use the parabola's equation)

What then is the area of the rectangle? (Hint: Multiply height by width)

Then what is the total area of the whole rectangle? (Hint: Multiply by 2)

 

[blamocur]

y=5−x 2 .

I am guessing this means [imath]y = 5-x^2[/imath].

 

[Akalan]

I am guessing this means [imath]y = 5-x^2[/imath].

yes thats right

 

[Steven G]

A similar was posted recently and I'll give the same as advice as I gave the other poster.
If you pick any point on the parabola (other than the max point) and using this as a point on the rectangle, then this point will define the rectangle. Think about how you will draw the inscribed rectangle with just one point. Call this point (x,y) noting that y= 5-x^2. What is the length and with of this rectangle (you certainly can use staples hint about just using half the rectangle)? What is the maximum area.