Expert: Paul Klarreich - 9/24/2008

Question
1.a piece of wire16 in. long is cut into two pieces; 1 piece is to be bent to form a circle and the other piece is to be bent to form a square. Where is the cut made if the sum of the areas of the square and the circle is a minimum.

2. Find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area.

3. a window is made in the shape of a rectangle surmounted by a semi-circle whose diameter is equal to the width of the rectangle. If the perimeter of the window is 16ft, what dimension will admit the most light?

Answer
Questioner:   monica
Category:  Calculus
Private:  No
 
Subject:  help!!
Question:  1.a piece of wire16 in. long is cut into two pieces; 1 piece is to be bent to form a circle and the other piece is to be bent to form a square. Where is the cut made if the sum of the areas of the square and the circle is a minimum.

2. Find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area.

3. a window is made in the shape of a rectangle surmounted by a semi-circle whose diameter is equal to the width of the rectangle. If the perimeter of the window is 16ft, what dimension will admit the most light?
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Hi, Monica,

Is this your first attempt at Max-min problems?  If so, the scheme is something like this:

1. Identify the variable(s) in the problem -- the things that can be changed.  Give them names.

2. Write the thing to be Max-ed(or min-ed) in terms of the variable(s).

3. If there are more than one, determine a relationship between the variables (called a 'CONSTRAINT') that will eliminate all but one.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is probably the hardest part.

4. Now differentiate, set the derivative = 0, and solve.  Be sure to check for endpoint max-mins.

AND, Please check the archives for other M-M examples.  There are a lot of them.  Click BROWSE PAST ANSWERS.
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Since this appears to be your whole homework assignment, I'll try to get you started on all of them, and you can finish up.

1.a piece of wire  16 in. long is cut into two pieces; 1 piece is to be bent to form a circle and the other piece is to be bent to form a square. Where is the cut made if the sum of the areas of the square and the circle is a minimum.

Let x = the piece of wire that will make the circle.

THEN

16-x is the piece for the square.

4 - x/4 is a side of the square.

(4 - x/4)^2 is the area of the square

x/(2pi) is the radius of the circle.

pi (x/2pi)^2 is the area of the circle.
(pi/4)x^2  is the area of the circle.


A = (4 - x/4)^2 + (pi/4)x^2
is the area to be minimized.  

CHECK THE ARCHIVES.

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2. Find the equation of the line through the point (3,4) which cuts from the first quadrant a triangle of minimum area.

The equation has the form:  y - 4 = m(x - 3). (and clearly m < 0)

Find the intercepts in terms of  m:

The x-intercept:

0 - 4 = m(x0 - 3)

- 4/m = x0 - 3

x0 = 3 - 4/m

The y-intercept:

y0 - 4 = m(0 - 3)

y0 - 4 = -3m

y0 = 4 - 3m

Now the area of the triangle is:

A = x0y0/2

You can handle the rest.

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3. a window is made in the shape of a rectangle surmounted by a semi-circle whose diameter is equal to the width of the rectangle. If the perimeter of the window is 16ft, what dimension will admit the most light?

Recognize this?  It's pretty much the same as #1.

Let w = the width of the rectangle, AND the diameter of the semicircle.
Let h = the height of the rectangle.

Then the circumference of the semicircle is pi w/2.

and the area of the semicircle is 1/2 pi (w/2)^2 = piw^3/8

The total area is A = piw^3/8 + wh.

We need to eliminate a variable.  Use our constraint:

pi w/2 + w + 2h = 16

Solve for h:

pi w/2 + w - 16 = 2h

h = pi w/4 + w/2 - 8

A = piw^3/8 + w(pi w/4 + w/2 - 8).

Simplify, get A', set = 0, solve.