The Lagrangian is
It has critical points where the first order derivatives vanish:
From the first two equations we get
Then
At these critical points, we have
(maximum)
(minimum)
For this, we use simultaneous equations. Let George's page be g, Charlie's be c and Bill's page be b.
First, <span>George's page contains twice as many type words as Bill's.
Thus, g = 2b.
</span><span>Second, Bill's page contains 50 fewer words than Charlie's page.
Thus, b = c - 50.
</span>If each person can type 60 words per minute, after one minute (i.e. when 60 more words have been typed) <span>the difference between twice the number of words on bills page and the number of words on Charlie's page is 210.
We can express that as 2b - c = 210.
Now we need to find b, since it represents Bill's page.
We can substitute b for (c - 50) since b = c - 50, into the equation 2b - c = 210. This makes it 2(c - 50) - c = 210.
We can expand this to 2c - 100 - c = 210.
We can simplify this to c - 100 = 210.
Add 100 to both sides.
c - 100 + 100 = 210 + 100
Then simplify: c = 210 + 100 = 310.
Now that we know c, we can use the first equation to find b.
b = c - 50 = 310 - 50 = 260.
260 is your answer. I don't know where George comes into it. Maybe it's a red herring!</span>
The speed of the ball is
ds/dt = 32t
At t =1/2 s
ds/dt = 16 ft/s
The distance from the ground
50 - 16(1/2)^2 = 46 ft
The triangles formed are similar
50/46 = (30 + x)/x
x = 345 ft
50 / (50 - s) = (30 + x)/x
Taking the derivative and substituing
ds/dt = 16
and
Solve for dx/dt
Answer:
-1
Step-by-step explanation:
Has to be the opposite
