Man

[PrudensOptimus]

A man 1.75 m tall walks at a rate of 2m/s toward a streetlight that is 10m above the ground, at what rate is the length of the shadow changing when he is 6m from the base of the light?


Don't know what to do. => Pythagorean?


#2: A rectangle with base an x axis is inclined under y = 6 - x^2, Find the height of the rectangle whose area is as large as possible.

h = 6, when y' = 0?

A = 6*base, A' = 6?

[StephenPrivitera]

If I understand #2 correctly
Consider only x positive.
A=bh=(2x)(y)=(2x)(6-x2)=12x-2x3
dA/dx=12-6x2=0
12=6x2
x=sqrt2
This is the only solution (if x were negative, you'd get the same area) so it's probably a max. Check just to be sure.
d2A/dx2=-12x<0
yup a max
(by the way x=-sqrt2 appears to be a min, but that's because my original equation should be A=2|x|y)
So the height is y=6-2=4

[himanshu121]

For the first part draw the diagram relate the heights with distance from the post By applying Similar Triangle Prop
.

[HallsofIvy]

In 1, you do not use the Pythagorean theorem because you are not given and don't want to find, the length of the hypotenuse. You can, as himanshu said, use similar triangles. The triangle formed by the tip of the man's shadow, the base of the lightpost, and the top of the lightpost is similar to the triangle formed by the tip of the man's shadow, the man's feet, and the man's head.

In 2, I was a bit confused by the word "inclined"! I assume you mean that the rectangle has two vertices on the x-axis and two vertices on the parabola. The two vertices on the parabola have coordinates (x, 6- x2) so the base of the rectangle has length 2x and the height is 6-x2. The area of the rectangle is, then, 2x(6-x2)= 12x- 2x3, just you have.

[PrudensOptimus]

Originally posted by himanshu121
For the first part draw the diagram relate the heights with distance from the post By applying Similar Triangle Prop
.

I tried x/y = x1/y1, but... i don't know what did the book mean by the guy's shadow...

[himanshu121]

Connect the topmost point of lamp post and man's head extend it till it reaches the ground you may draw the shadow by the laws of optics

The distance b/w the feet of man and the point u get by extending is the length of the shadow