Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 19

MATH100 A December 2023

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Question 19
Rectangle inscribed in a parabola A rectangle is inscribed with its base on the $x$-axis and its upper corners on the parabola $y=4-x^{2}$, as shown. What is the height of such a rectangle with the greatest possible area?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
This is an optimization problem. Can you write down a 1-variable formula for the area of a rectangle that is inscribed in the region between the parabola and $x$-axis?

Hint 2

Let $R$ be a rectangle that is inscribed in the region between the $x$-axis and the parabola. We may as well parametrize the rectangle by the $x$-coordinate of its bottom-right corner. Let us call this coordinate $x$. Then, the top-right corner of the rectangle lies on the parabola, so its $y$-coordinate is $4-x^{2}$. Thus, the area of the rectangle is given by $A(x)={\text{base}}(x)\cdot {\text{height}}(x)=(2x)\cdot (4-x^{2})=2(4x-x^{3}).$

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We optimize the area function $A$ from the hint above. First, its derivative is given by $A'(x)=2(4-3x^{2}).$ Setting $A'(x)=0$, we find $4-3x^{2}=0$, which holds for $x={\frac {\pm 2}{\sqrt {3}}}$. Since $x$ is the coordinate of the bottom-right corner, it is positive, so the only extremum of $A$ occurs at $x={\frac {2}{\sqrt {3}}}$. To perform our due diligence, we should check that this is indeed a maximum by computing the value $A''\left({\frac {2}{\sqrt {3}}}\right)$. We have $A''(x)=2(-6x),$ so $A''\left({\frac {2}{\sqrt {3}}}\right)<0$. Thus, the rectangle with maximal area has its bottom-right corner at the point $\left({\frac {2}{\sqrt {3}}},0\right)$. The height of this rectangle is $4-\left({\frac {2}{\sqrt {3}}}\right)^{2}=4-{\frac {4}{3}}={\frac {8}{3}}$.