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Max/Min Applications of Derivatives

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Date: 05/31/98 at 20:35:41
From: Beth
Subject: Applications of the Derivative

1. If 40 passengers hire a train, it will cost \$8.00 per person, but
the fare will be cut by \$0.10 apiece to for every passenger over
40. What number of passengers will make the most income for the

2. Find the dimensions of the rectangle of largest area that can be
inscribed in an isosceles right triangle, with 2 sides of the
rectangle lying along the triangle's legs.

3. How should a number be divided into a sum so that the product of 2
summands is as large as possible?

4. A 10-foot sign is 10 ft. off the ground. Find the distance from the
sign of a point on the ground from which the subtended angle of the
sign is a maximum.

I looked through all of my calculus and trig notes and could not find
how to do these problems. Your help would be greatly appreciated.
```

```
Date: 06/01/98 at 07:40:49
From: Doctor Anthony
Subject: Re: Applications of the Derivative

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Problem 1:

If n is the number of passengers over 40, then the total income of the

T = (40+n)[8 - 0.1n]  = 320 - 4n + 8n - 0.1n^2

T = 320 + 4n - 0.1n^2

dT/dn = 4 - 0.2n = 0  for maximum

0.2n = 4

n = 4/0.2 = 20

So the optimum number of passengers is 40 + 20 =  60.

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Problem 2:

Take the origin at the right angle and let the sides of the rectangle
have lengths x and y.

Now the coordinates of the corner of the rectangle opposite the origin
are (x,y), the corner lies on the hypotenuse of the triangle, and
this line has equation x + y = L where L is the length of one of the
legs of the isosceles triangle.

Putting y = L-x, the area of the rectangle is xy = x(L-x):

A = Lx - x^2

dA/dx = L - 2x = 0 for max or min

2x = L

x = L/2

Then y = L-x = L/2.

So the largest rectangle is a square of side L/2.

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Problem 3:

Let N be the number and suppose x+y = N so y = N-x.

Maximize the product xy = x(N-x):

P = Nx - x^2

dP/dx = N - 2x = 0 for maximum.

2x = N

x = N/2  and  y = N/2

So you should divide the number in half to get a maximum product.

---------
Problem 4:

Let x be the distance of the point on the ground from the vertical
through the sign.

Let  A = angle of elevation from point to top of the sign
B = angle of elevation from point to bottom of the sign

Tan(A) = 20/x and tan(B) = 10/x

tan(A) - tan(B)     20/x - 10/x    20x - 10x
tan(A-B) = ------------------ = ------------- = ---------
1 + tan(A)tan(B)    1 + 200/x^2    x^2 + 200

10x
tan(A-B) =  ---------
x^2 + 200

So we require to maximize this expression in x.

(x^2 +200)10 - 10x(2x)
d tan(A-B)/dx =  ----------------------  = 0 for max or min
x^2 + 200

Thus we require the numerator to be zero:

10(x^2 + 200) - 20x^2 = 0

-10x^2 + 2000 = 0

x^2 = 200

x = 14.14 ft.

So the point on the ground should be 14.14 feet from the poster stand.

-Doctor Anthony,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/
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Associated Topics:
High School Calculus

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