Solving an Optimisation Problem

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.

• "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. Commented Jun 8, 2020 at 23:55

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 $$0, so that the rectangle wouldn't be 'empty' (or a line). If $$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, $$2(2a + 4-f(a))$$ $$4a + 4 - 2f(a)$$

• Why is the width $4-f(a)$ instead of $f(a)-4$? Commented Jun 9, 2020 at 0:01
• $|f(a) - 4|$ 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 $-2<a<2$ Commented Jun 9, 2020 at 0:04

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

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