Line Cutting off Triangle of Least Area

[turophile]

The problem:

Which line tangent to y = 9 – x[sup:19629od8]2[/sup:19629od8] cuts off the smallest area in the first quadrant?

My work so far:

Let the point of tangency be (h, 9 – h[sup:19629od8]2[/sup:19629od8]), let the x-intercept be (u, 0), and let the y-intercept be (0, v).
The area of the triangle is a = (1/2)uv. We need to express a in terms of h so we can find the tangent point that gives the smallest area for the cut off triangle.

The equation of the line is y – (9 – h[sup:19629od8]2[/sup:19629od8]) = m(x – h) or y = m(x – h) + 9 – h[sup:19629od8]2[/sup:19629od8]. The slope m is y' = – 2x. So the equation of the line is:

y = (– 2x)(x – h) + 9 – h[sup:19629od8]2[/sup:19629od8] = – 2x[sup:19629od8]2[/sup:19629od8] + 2xh + 9 – h[sup:19629od8]2[/sup:19629od8].

Then:

0 = – 2u[sup:19629od8]2[/sup:19629od8] + 2uh + 9 – h[sup:19629od8]2[/sup:19629od8]
? 2u[sup:19629od8]2[/sup:19629od8] – 2uh = 9 – h[sup:19629od8]2[/sup:19629od8]
? u[sup:19629od8]2[/sup:19629od8] – uh = (9 – h[sup:19629od8]2[/sup:19629od8])/2
? u[sup:19629od8]2[/sup:19629od8] – uh + h[sup:19629od8]2[/sup:19629od8]/4 = (u – h/2)[sup:19629od8]2[/sup:19629od8] = (9 – h[sup:19629od8]2[/sup:19629od8])/2 + h[sup:19629od8]2[/sup:19629od8]/4
? u = sqrt[(9 – h[sup:19629od8]2[/sup:19629od8])/2 + h[sup:19629od8]2[/sup:19629od8]/4] + h/2 = (1/2) sqrt[18 – 2h[sup:19629od8]2[/sup:19629od8] + h[sup:19629od8]2[/sup:19629od8]] + h/2 = (1/2) sqrt[18 – h[sup:19629od8]2[/sup:19629od8]] + h/2

and:

v = 9 - h[sup:19629od8]2[/sup:19629od8]

So:

a = (1/2) ? {(1/2) sqrt[18 – h[sup:19629od8]2[/sup:19629od8]] + h/2} ? {9 - h[sup:19629od8]2[/sup:19629od8]}

Does this look correct so far? I've been trying to calculate the derivative of this with respect to h, but haven't been able to get the answer in my textbook, which is y + 2 sqrt(3) x = 12.

 

[galactus]

Re: Line Cutting of Triangle of Least Area

The area of the triangle is \(\displaystyle A=\frac{xy}{2}\)

Plug in the given y equation and get \(\displaystyle A(x)=\frac{1}{2}x(9-x^{2})\)

Differentiate: \(\displaystyle A'(x)=\frac{-3}{2}x^{2}+\frac{9}{2}\)

Set to 0 and solve for x gives us \(\displaystyle x=\pm\sqrt{3}\)

Since we are in the first quadrant we go with the positive one.

The derivative of the parabola is \(\displaystyle -2x\)

This means at \(\displaystyle x=\sqrt{3}\), the slope is \(\displaystyle m=-2\sqrt{3}\)

And at \(\displaystyle x=\sqrt{3}\), the y value is \(\displaystyle 9-(\sqrt{3})^{2}=6\)

Now, use \(\displaystyle y=mx+b\) and solve for b.

\(\displaystyle 6=-2\sqrt{3}(\sqrt{3})+b\Rightarrow b=12\)

So, the line equation is \(\displaystyle y=-2\sqrt{3}x+12\) as given.

 

[turophile]

Re: Line Cutting of Triangle of Least Area

Thanks! I see where the answer comes from now. But I'm confused about how A = (xy)/2. Isn't that half of the area of the rectangle with vertices (0, 0), (x, 0), (0, y), and (x, y), which is less than the area in the first quadrant under the parabola, which is in turn less than the area of the triangle cut off by the line tangent to the parabola at (x, y)? Maybe I'm visualizing the triangle the wrong way?

 

[galactus]

Here is a graph of what we are talking about.

xy/2 is the area of a triangle.

The triangle has area \(\displaystyle 12\sqrt{3}\approx 20.78\)

The area under the parabola in the first quadrant has area 18

See, the triangle has an area slightly larger than the area under the parabola.

EDIT: I made an error in the area under the parabola. I fixed it.

 

[turophile]

Thanks for the graph. That's what I was visualizing. So if A = xy/2, it seems to me that x is the value of the x-intercept and y is the value of the y-intercept of the tangent line. But if y = 9 - x[sup:1ppwidkw]2[/sup:1ppwidkw], as you showed in your calculations, doesn't that also mean that (x, y) is the point on graph where the tangent line touches? Can the same x and y values work in both contexts? Or am I confusing the referents of x and y in your explanation? Somehow defining the area as A = xy/2 makes me think of the area of a triangle inscribed in the parabola, not a triangle that contains the parabola. But I see how your calculations get the answer in the textbook. I'll think about it some more and hope that something clicks.