Calculus rate of change problem?

Question

Could someone solve this calculus problem with detailed work?

Find the length of the longest rod which can be carried horizontally (level) around a corner from a corridor 8 feet wide into one 4 feet wide.

DO NOT USE TRIGONOMETRY.

Thanks.

Answer 1

OK, sketch the corner so the 8 ft corridor is horizontal and the 4 ft corridor is vertical. Draw the rod so it touches outer walls and inside corner, making a right triangle with the outer walls. Extend inner walls (dotted lines). Top triangle has a base of 4; let's call height y (the amount the rod extends vertically beyond the turn). Bottom triangle has height 8, call base x (amount rod extends horizontally beyond turn). The 2 triangles are similar, so y/4 = 8/x, y = 32/x. Let L be length of rod.

L² = (x+4)² + (y+8)²
L² = x² + 8x + 16 + y² + 16y + 64
L² = x² + 8x + 16 + (32/x)² + 16(32/x) +
64
L² = x² + 8x + (32/x)² + 16(32/x) +
80

We need to minimize L². (We could take square root and minimize L, but when we find derivative, that puts 2√(L²) on the bottom of fraction, and then we multiply it out again). So

(L²)' = 2x + 8 - 512/x² - 2048/x^3 = 0

x^4 + 4x^3 - 256x - 1024 = 0
x^3(x+4) - 256(x+4) = 0
x^3 = 256
x = cube root of 256 = 4 cube root of 4.
x ≈ 6.35
y = 32/x ≈ 5.04

L = √[ (6.35+4)² + (5.04+8)²] ≈16.65

So rod length must be < 16.65.

Answer 2

I presume it is the right angle at the corner.
The length of the rod is:
l = sqrt(64 + y^2) + sqrt(x^2 + 16)..........(1), where
8/x = y/4 therefore xy = 32...........(2).
For maximum l we must have:
dl/dx = x/(sqrt(x^2 + 16) - (32/x^2)* y/sqrt(y^2 + 64) = 0..........(3)
Using (2) y/sqrt(y^2 + 64) = 4/sqrt(x^2 + 16) and (3) yields:
x = 4*2^(1/3) and (with (2)) y = 8/2^(1/3)..............(4)
You can verify that the second derivative of l at this point equals 3 > 0, therefore we have got minimal l.
Plugging (4) into (1) we obtain maximum (that will pass which is our minimal l) rod length:
lmax = 4*(1 + 2^(1/3))*sqrt(1 + 2^(2/3)) = 14.540694 feet.

Cheers.

Answer 3

I'm sorry. I gave you the wrong answer.
I thought I could solve it using pure geometry.

What I'm going to do requires the math knowledge of a 11-th grade pupil ore above.

suppose the angles are right.
we have y/8 = x/4 ==> xy = 32;

We need to minimize L = sqrt((x+4)^2+(y+8)^2) subject to xy = 32

L = sqrt((x+4)^2 + (32/x+8)^2)
As usual take derivative of L with respect to x; and solve to find its zero;
It is minimized at x = 4 2^(2/3)

Resubstitute we have L ~= 16.65 feet.

The maximum length of the rod should not exceed this number. Therefore, the maximum length is also 16.65 feet

Answer 4

I get the maximum length to be 16.65ft approx. It was done using differentiation on trigonometric functions. I don't see how you can do it any other way.
You want to find the minimum of 4cscA + 8secA. This occurs when tanA = 2^(-1/3) and then gives the result above.

Answer 5

I would just solve in terms of y, really not sure how one would use calculus to do this problem. That being said I would just make 240 mph the hypotenuse, then take y = 240*Sin(22degrees) =89.91mph. The airplane is gaining altitude at 89.91 mph.