calculus-optimization problem?

Question

A right triangle in the first quadrant has the coordinate axes as sides, and the hypotenuse passes through the point (1,8). Find the vertices of the triangle so that the length of the hypotenuse is a minimum.

Answer 1

Write an equation of the line through (1, 8). The slope m is what we're trying to find:

m = (y - 8)/(x - 1)
m(x - 1) = y - 8

Find the y-intercept where x = 0:

-m + 8 = y

Find the x-intercept where y = 0:

x = 1 - 8/m

The length of hypotenuse would be:

sqrt((1 - 8/m)^2 + (8 - m)^2)

But it's easier just to work with the square of the hypotenuse to avoid the chain rule.

Expand:

1 - 16/m + 64/m^2 + 64 - 16m + m^2

Collect terms:

m^2 - 16m + 1 - 16m^(-1) + 64m^(-2)

Differentiate with respect to m:

2m - 16 + 16m^(-2) - 128m^(-3) = 0

Multiply through by m^3 and divide by 2:

m^4 - 8m^3 + 8m - 64 = 0

m^3(m - 8) + 8(m - 8) = 0
(m^3 + 8)(m - 8) = 0
(m + 2)(m - 8)(m^2 + 2m + 4) = 0

A positive slope wouldn't result in a triangle, so m = -2.

Then y = -(-2) + 8 = 10 when x = 0, and x = 1 + 8/2 = 1 + 4 = 5 when y = 0.

So I think it's (0, 10) and (5, 0) and the origin (0, 0).

(I always screw something up the first time...)

----

I thought ironduke8's answer might be what I'd get, but just to check... The length of his hypotenuse would be 9√2 ≈ 12.73.

The length of mine and Puggy's would be √125 = 5√5 ≈ 11.18.

We win. :c)

Answer 2

We know that the equation of a line is y = mx + b.
Therefore, since we're given the point (1, 8), we have the equation

8 = m(1) + b
8 = m + b

Since the coordinate axes are the sides, we need to determine the x and y intercepts of the unknown line y = mx + b.

x-int: make y = 0. Then 0 = mx + b and x = -b/m
y-int: This is just b.

Thus, our vertices are located at:

(-b/m, 0) and (0, b).

What we want to do is minimize the distance between them. But we have a distance formula;

d = sqrt ( (y2 - y1)^2 + (x2 - x1)^2 )

Plugging in our values (-b/m, 0) and (0, b) as (x1, y1) and (x2, y2), we have

d = sqrt ( (b - 0)^2 + (0 - (-b/m))^2 )
d = sqrt ( b^2 + (b/m)^2 )
d = sqrt ( b^2 + (b^2)/(m^2) )

Let's put this under a common denominator inside the square root.

d = sqrt ( [(b^2)(m^2) + (b^2)] / (m^2) )

And now, let's pull out b^2/m^2 from the square root. Taking it out of the square root will result in it becoming b/m, and we're left with

d = (b/m) sqrt (m^2 + 1)

However, we have the relationship that m + b = 8, or
b = 8 - m. Let's plug that in.

d = (8 - m)/m sqrt (m^2 + 1)

This will be our distance function, d(m).

d(m) = [(8 - m)/m] sqrt (m^2 + 1)

Now, we solve for the derivative and make it 0. We know that we're going to eventually make d'(m) equal to 0, so let's square both sides.

[d(m)]^2 = (8 - m)^2 / m^2 [m^2 + 1]

[d(m)]^2 = (64 - 16m + m^2)(m^2 + 1) / m^2

[d(m)]^2 = [64m^2 + 64 - 16m^3 - 16m + m^4 + m^2] / m^2

[d(m)]^2 = (m^4 - 16m^3 + 65m^2 - 16m + 64) / (m^2)

[d(m)]^2 = m^2 - 16m + 65 - 16/m + 64/m^2

Differentiating implicitly, we get

2[d(m)] [d'(m)] = 2m - 16 + 16/m^2 - 128/m^3

Setting d'(m) to 0,

0 = 2m - 16 + 16/m^2 - 128/m^3

Now, we solve for this.

128/m^3 - 16/m^2 = 2m - 16

Multiply both sides by m^3,

128 - 16m = 2m^4 - 16m^3

Moving everything to the right hand side,

0 = 2m^4 - 16m^3 + 16m - 128

Dividing by 2,

0 = m^4 - 8m^3 + 8m - 64

This actually groups nicely,

0 = m^3 (m - 8) + 8(m - 8)
0 = (m^3 + 8) (m - 8)
0 = (m + 2) (m^2 - 2m + 4) (m - 8)

Therefore, our critical values are m = {-2, 8}.

Since our solution lies in the first quadrant, we know our slope has to be downward; m = -2.

If m = -2, then b = 10.

Our base is -b/m and our height is b, so we have
-10/-2 = 5 as our base, and 10 as our height.

Therefore, the vertices that make the length of the hypotenuse a minimum would be (5, 0) and (0, 10), and, of course, the origin (0,0).

Remember that the line joining these two would be given as

y = -2x + 10, so it would pass through the point (1,8).

Answer 3

The hypotenuse will have a minimum length when m = -1
So y=-x + b
8 = -1 +b
b=9
y = -x + 9
x- intercept = (9,0)
y- intercept = (0,9)
origin = (0,0)
These are the three vertices of the right triangle.