Help please!

[tabitha.halford]

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

How would I set up a problem to find the solution for this?​

 

[Bob Brown MSEE]

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

How would I set up a problem to find the solution for this?​

y = 2- 5x^2 would make more sense.
Otherwise the area is unbounded.

Check to see if the problem was written this way.
Thanks

 

[mmm4444bot]

Please check that you correctly posted the quadratic equation:

y = 5x^2 - 2

It is not possible to inscribe a rectangle (with base on x-axis) within that parabola because it opens upward.

We need a parabola that opens downward with y-intercept greater than zero.

Like, for example:

 

[tabitha.halford]

Sorry, I made a type error. The problem is y = 5-x²

 

[mmm4444bot]

:idea: Okay -- there is a [Preview Post] button next to the submit button; you may use it to proofread your posts before submission.


Do you see in my graph that the rectangle is horizontally centered? Half of the rectangle lies to the right of y-axis, and the other half lies to the left of y-axis. Your representative, inscribed rectangle will be the same -- but the parabola will look different because the quadratic is:

y = -x^2 + 5

So, we see that symbol x represents only half of the base.

Base of rectangle is 2x

Height of your rectangle is the value of y (do you see why it's y?)

Height of rectangle is -x^2 + 5


You know what the area formula for a rectangle is, yes?


Edit: Deleted nonsense about finding vertex of new quadratic polynomial, after reading posts by Bob & Jeff. You may use calculus or graphing technology, to find maximum value of cubic polynomial within appropriate domain.


Please show us how far you get now. Cheers :cool:

 

[Bob Brown MSEE]

I am suprised to see this listed as an Advanced Algebra question.
Do you know how to find the local maximum of a cubic on 0<x<2.24?
You may need to use a graphing calculator.

 

[JeffM]

Hi Tabitha

In one of your previous posts, you said you used derivatives. That means you have some knowledge of differential calculus.

99% of the practical use of differential calculus is finding local maxima or minima (more generally local extrema). This problem is a maximization problem so you need to apply your knowledge of derivatives.

So are you good to go now?

 

[mmm4444bot]

I am suprised to see this listed as an Advanced Algebra question.

You may need to use a graphing calculator.

Good call, Bob (and Jeff). We see these types of exercises from algebra courses, occasionally. Students are instructed to use zooming or some technology, to obtain final solution, but they usually neglect to include that instruction in their post.

Not sure what this student was told.

(Also not sure whether my latest wave of dementia is the beginning of a new chapter in my life!)

 

[JeffM]

Don't worry, Mark. It is so easy to make a misstep in postings where 80% of the challenge is trying to figure out what the actual problem is. I do it frequently.