0
$\begingroup$

A rectangle with sides parallel to the coordinate axes is inscribed in the region enclosed by the graphs of $y=x^2$ and $y=4$.

The first part of the question is to sketch the graph and show the region that is under consideration, which I have done. Although, I'm quite unsure if I have answered it properly as I do not quite fully understand the sentence: "...with sides parallel to the coordinate axes..."

What I'm lost on what to do next is the second part of the question, it states:

Supposing that the x-coordinate of the bottom-right vertex of the rectangle is a. Specify the possible values of a and find a formula which expresses the length of the perimeter P as a function of a.

$\endgroup$
1
  • $\begingroup$ "Parallel to the axes" just means that two of the sides are horizontal (parallel to the $x$-axis) and two are vertical (parallel to the $y$-axis.) "Inscribed in the parabola" means that two of the vertices of the rectangle are on the line, and two are on the parabola. $\endgroup$
    – saulspatz
    Commented Jun 8, 2020 at 23:55

2 Answers 2

3
$\begingroup$

Firstly, this graph looks like a parabola chopped off at $ y = 4 $. The possible values of $a$ on the bottom right would then be $0\le a \le 2 $, since $$f(x) = x^2$$ $$f(a) = 4$$ $$a = 2$$ However, I would say that the more 'correct' answer is $0<a<2$, so that the rectangle wouldn't be 'empty' (or a line). If $ a = 0$, that means its just a vertical line on the y-axis. If $a = 2$, its a horizontal line on $y = 4$.

The perimeter can be separated into 2 parts, width and length. The length would be simply $a - (-a) = 2a$, since it is the different of co-ordinate of x on the left and the right side.

The width would be the difference between the horizontal line on $y = 4$ and whatever the value of $y$ is on $x = a$. Thus , width = $4 - f(a)$

The perimeter would be twice the sum of width and length, $$ 2(2a + 4-f(a)) $$ $$ 4a + 4 - 2f(a) $$

$\endgroup$
2
  • $\begingroup$ Why is the width $4-f(a)$ instead of $f(a)-4$? $\endgroup$
    – RobiChilla
    Commented Jun 9, 2020 at 0:01
  • $\begingroup$ $|f(a) - 4|$ also works, since we're looking for distance which is always positive. Howeer I wrote $ 4 - f(a) $ so that the answer is already positive. $y = 4$ is the upper bound (located on the top), and will always be above $f(a)$ as long as $-2<a<2$ $\endgroup$ Commented Jun 9, 2020 at 0:04
1
$\begingroup$

The above answered the optimization problem. Here is a picture (since you asked) parabola rectangle

$\endgroup$
2
  • $\begingroup$ Ignore the green line below the parabola. Couldn't figure out how to get rid of it. $\endgroup$
    – K.defaoite
    Commented Jun 8, 2020 at 23:53
  • $\begingroup$ Thank you, I was thinking of a totally different way of putting in a rectangle in there hahaha. This is really helpful, thank you. $\endgroup$
    – RobiChilla
    Commented Jun 8, 2020 at 23:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .