Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 28(b)

MATH100 A December 2023

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[hide] Question 28(b)

Let $$f(x)=e^{-x}.$$ The goal of this question is to find the number ${\textstyle a}$ such that the triangle consisting of the portion of the first quadrant that lies below the tangent line to ${\textstyle f(x)}$ at ${\textstyle x=a}$ has the largest possible area. Part B: Find the number ${\textstyle a}$ such that the triangle consisting of the portion of the first quadrant that lies below the tangent line to ${\textstyle f(x)}$ at ${\textstyle x=a}$ has the largest area possible.

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.

[show] Hint 1
From the plot, you can see that the triangle has vertices at the origin, the ${\textstyle y}$-intercept of the tangent line, and the ${\textstyle x}$-intercept of the tangent line. Can you make an expression for the area of the triangle that depends on ${\textstyle a}$?

[show] Hint 2
This is a maximisation problem, you will need to construct an expression for the area of the triangle in terms of the point ${\textstyle x=a}$. First determine how the ${\textstyle x}$- and ${\textstyle y}$-intercepts depend on the point ${\textstyle x=a}$, then use the formula for the area of the triangle.

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[show] 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. The first step should be setting up an equation for the area of the triangle that is in terms of the point ${\textstyle a}$. The slope of the tangent line at $x$ is ${\textstyle f'(x)=-e^{-x}}$, so the equation of the tangent line at $x=a$, using the point-slope formula to start, is: $${\begin{aligned}(y-e^{-a})&=-e^{-a}(x-a),\\y&=-e^{-a}x+e^{-a}(1+a).\end{aligned}}$$ We can read off the ${\textstyle y}$-intercept as ${\textstyle y=e^{-a}(1+a)}$, and a simple rearranging will show that the ${\textstyle x}$-intercept is ${\textstyle x=1+a}$. With this, we now know the height and base of the right angle triangle whose area $A$ we want to find. This is: $$A(a)={\frac {1}{2}}(1+a)(e^{-a}(1+a))={\frac {1}{2}}e^{-a}(1+a)^{2}$$ Note that if ${\textstyle a<-1}$, then the ${\textstyle x}$-intercept is less than zero, and there would be no area in the first quadrant. Because of this, we restrict ${\textstyle a\geq -1}$. Now, to find the optimal ${\textstyle a}$ we need to calculate ${\textstyle A'(a)}$ and solve for the critical points. $${\begin{aligned}A'(a)=e^{-a}(a+1)-{\frac {1}{2}}e^{-a}(1+a)^{2}\end{aligned}}$$ Setting ${\textstyle A'(a)=0}$ we have: $${\begin{aligned}e^{-a}(a+1)-{\frac {1}{2}}e^{-a}(1+a)^{2}&=0\\e^{-a}\left(a+1-{\frac {1}{2}}(1+2a+a^{2})\right)&=0\\{\frac {1}{2}}a^{2}&={\frac {1}{2}}\\a&=\pm 1\end{aligned}}$$ And since we know that if ${\textstyle a=-1}$, then the area in the positive quadrant is $0$, it follows that the point ${\textstyle a}$ which maximises the area is ${\textstyle a=1}$.