Maximize area of triangle in the Cartesian coordinate system

calculus

Question:

Consider the function

$$\large f(x) = xe^{-\frac{x}{2}}$$

Let $P$ be a point on the curve $y = f(x)$ with the $x$-coordinate $a$ (where $a > 2$).

Let $R$ be the point where the normal line in $P$ cuts the x-axis.

Let $Q$ be the point $(a,0)$

Investigate if there is any value of $a$ for which the area of the triangle $PQR$ is maximum and, if this is the case, determine $a$.

The solution must use differentiation and optimization, so no smart non-calculus solution are allowed.

Attempted solution:

Let us draw an image. It would look something like this:

enter image description here
Don't pay attention to the x-coordinate in the graph for P, Q and R. Those are not necessarily the true values but just the values those points got due to quickly drawing the image freehand.

Here is my basic strategy:

  1. Write down the coordinates for P.
  2. Take the derivative of f(x).
  3. Find the k value of the normal.
  4. Use the point-slope equation to get an expression of b in terms of a.
  5. Find the base and height of the triangle.
  6. Write down an expression for the area of the triangle.
  7. Take the derivative and find local maximum. This local maximum is the sought after $a$. Here I will make the risky assumption that the area is not maximum when a goes to infinity since I know the true value of a.

1. Write down the coordinates for P.

Basically, substitute x for a in the f(x) equation:

$$\large P = (a, ae^{-\frac{a}{2}})$$

2. Take the derivative of f(x).

Here I use the product rule and get:

$$\large f'(x) = 1 \cdot e^{-\frac{x}{2}} + x(-\frac{1}{2} e^{-\frac{x}{2}}) = e^{-\frac{x}{2}}(1-a) = \frac{(1-a)}{e^{\frac{x}{2}}}$$

3. Find the k value of the normal.

Here I make use of the formula that the product of the k for the tangent line and normal line equals to minus 1.

$$\large k_t k_n = -1 \Rightarrow k_n = -\frac{e^{\frac{a}{2}}}{1-a}$$

  1. Use the point-slope equation to get an expression of b in terms of a.

I am using the points P and R here:

$$\large (ae^{\frac{a}{2}} – 0) = -\frac{e^{\frac{a}{2}}}{(1-a)} (a-b)$$

This produces the value b:

$$b = \frac{(1-a)a}{e^a} + a$$

5. Find the base and height of the triangle.

$$\large Base = a – b = a – (\frac{(1-a)a}{e^a} + a) = \frac{(a-1)a}{e^a}$$

$$\large Height = ae^{-\frac{a}{2}} – 0 = ae^{-\frac{a}{2}}$$

6. Write down an expression for the area of the triangle.

The area of a triangle is the base times the height divided by two.

$$\large Area = \frac{Bh}{2} = \frac{1}{2} \frac{(a-1)a}{e^a} \cdot ae^{-\frac{a}{2}} = \frac{a^3 e^{-\frac{3a}{2}}}{2} – \frac{a^2 e^{-\frac{3a}{2}}}{2}$$

7. Take the derivative and find local maximum.

Here I used Wolfram Alpha because of the complexity:

$$\large A'(a) = a(-0.75a^2 + 2.25a -1)e^{-\frac{3a}{2}}$$

Setting this to zero produces the following equation:

$$\large ae^{-\frac{3a}{2}} (-0.75a^2 + 2.25a -1) = 0$$

The three solutions from this is:

$$\large a_1 = 0$$
$$\large a_2 = \frac{3}{2} – \frac{\sqrt{\frac{11}{3}}}{2}$$
$$\large a_3 = \frac{3}{2} + \frac{\sqrt{\frac{11}{3}}}{2}$$

Since we know that $a >2$, we can neglect $a_1$ and $a_2$. So the solution then becomes:

$$\large a = \frac{3}{2} + \frac{\sqrt{\frac{11}{3}}}{2}$$

However, this is wrong as the expected answer is:

$$\large a = 2 + \frac{2}{\sqrt{3}}$$

Is the general approach I took reasonable? What went wrong? What are some productive ways to solve this problem and reach the expected answer that involves calculus-based optimization?

Best Answer

Area = 1/2·base·height

Let the base be the dist. between a and where the normal cuts X axis. This is called the length of subnormal of a fn at a pt(x0,y0) = (y0)·{f'(x0)}

So Area = (1/2)·(Base)·(Height)

Area = (1/2)·[f(a)·{f'(a)}]·[f(a)]

Substitute, differentiate to find maxima( if there)

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