Optimal fencing..?

Question

I don't know how to tackle these problems.
http://www.cet.ac.il/math/function/english/square/multi/multi7.htm
Can anyone help?

is there a derivative free way of doing this? i never knew algebra II required knowledge of derivatives.

Answer 1

In every problem like this, you will have two equations. One equation will be for the perimeter of the figure, and the other will be for the area. You will solve the perimeter equation for one variable and plug that into the area equation. You will then differentiate the area equation, set the derivative equal to zero, and solve. From this value you can calculate the other dimension.

For the first one, for example, let the "length" (the horizontal dimension up to where the circle starts) be l and the "width" (the vertical dimension) be w.

Then, for the perimeter equation, you have:

40 = 2l + w + πw/2 = 2l + w(1 + π/2)

The πw/2 arises because the circumeference of a full circle is π times the diameter, and the diameter of the half-circle on the right is w. But since it's only a half-circle, the distance around is π2/2.

The area would be the area of the rectangle (length times width or lw) plus the area of the half-circle (1/2 π (w/2)²) = π/8 w²).

Looks like it would be easier to solve the perimeter equation for l:

40 - w(1 + π/2) = 2l
2l = 40 - w(1 + π/2)
l = 20 - w/2(1 + π/2) = 20 - w(1/2 + π/4)

A = lw + π/8 w²

Substituting:

A(w) = (20 - w(1/2 + π/4))w + π/8 w²
A(w) = 20w - (1/2 + π/4 - π/8)w²
A(w) = 20w - (1/2 + π/8)w²
A(w) = 20w - (4 + π)/8 w²

Differentiate:

A'(w) = 20 - (4 + π)/4 w = 0
(4 + π)/4 w = 20
w = 80/(4 + π) ≈ 11.20

l = 20 - 80/(4 + π) * (2 + π)/4
l = 20 - 20(2 + π)/(4 + π) ≈ 5.6

That took a while...not 100% sure of my work...but that's the way you should attack them all.

Answer 2

For each problem, you want an equation for the perimeter (P) and an equation for the area (A). Labeling each picture first might help.

a) Let r = radius of the semicircle, w = width of the top part of the rectangle, h = height of the rectangle = 2r.

Then P = h + 2w + (1/2)*2*Pi*r = 2r + 2w + Pi*r
Also, A = wh + (1/2)*Pi*r^2 = 2rw+(1/2)*Pi*r^2

Since we know P = 40, we can substitute this into the first equation to get w in terms of r.
40 = 2r + 2w + Pi*r
2w = 40 - 2r - Pi*r
w = 20 - r - (1/2)*Pi*r

Plugging this w into the area equation we get:
A = 2r(20-r-(1/2)*Pi*r) + (1/2)*Pi*r^2
= -[(1/2)*Pi + 2]*r^2 + 40*r
Note: This can also be written as A = r(-[(1/2)*Pi + 2]*r + 40), a product of two linear functions.

Since we are trying to maximize the area, we will take the derivative of area with respect to r and then try to determine critical points:
A' = -[Pi+4]*r+40

Critical points occur when A' = 0:
-[Pi+4]*r+40=0
-[Pi+4]*r=-40
r = 40/[Pi+4] = 5.60 meters

Notice that A'' = -[Pi+4] < 0 always. So, the second derivative test gives us that r = 5.60 is the solution that maximizes area.

Substituting this value into the area equation, we get that the maximum area is:
A = -[(1/2)*Pi + 2]*(5.60)^2 + 40*(5.60) = 112.02 m^2

Hope that helps you for all the other parts.

Answer 3

1. You have 40 meters of fence, and you want to find the maximum area of a rectangle attached to a semicircle.

Let L be the length of the rectangle and W be the width. What we have are three sides of a rectangle and a circumference of a semicircle. Note that W also represents the diameter of the circle, so the perimeter is calculated as follows:

P = (top and bottom of rectangle) + (left side of rectangle) +
(circumference of the semicircle).

The top and bottom of the rectangle are obviously L. The left side of the rectangle is W, and the circumference of the semicircle is equal to half of the circumference of a circle.
The circumference of a circle is
C = pi(d)
But d = W, so C = pi(W). But we want HALF, so
(1/2)C = (1/2)pi(W). Therefore,

P = 2L + W + (1/2)pi(W).
But we know P = 40, so

40 = 2L + W + (1/2)pi(W). Multiplying both sides by 2,
80 = 4L + 2W + pi(W).

The area of the figure would be

A = (area of rectangle) + (area of semicircle)

Note that the area of the semicircle would be pi(r^2). However,
r = (W/2), so the area would be pi(W/2)^2, and we take half of this area for (1/2)pi(W/2)^2 = (1/2)pi(W^2/4) = (pi/8)(W^2)

A = LW + (pi/8)(W^2)

But, we can express L in terms of W in this equation:

80 = 4L + 2W + pi(W).
80 - 2W - pi(W) = 4L
L = [80 - 2W - pi(W)] / 4

A = {[80 - 2W - pi(W)] / 4}W + (pi/8)(W^2)
A = (80W - 2W^2 - pi(W)^2)/4 + (pi/8)(W^2)
A = [160W - 4W^2 - 2pi(W)^2 + pi(W^2)] / 8
A = [160W - 4W^2 - pi(W^2)] / 8
A = [160W - (4 + pi)W^2] / 8
A = (1/8) [160W - (4 + pi)W^2]

So this will be our area function, A(W).

A(W) = (1/8) [160W - (4 + pi)W^2]

To find the maximum area, we need to calculate the derivative with respect to W and then make it 0.

A'(W) = (1/8) [160 - 2(4 + pi)W]
A'(W) = (1/8) [160 - 2W(4 + pi)]

Making A'(W) = 0,
0 = (1/8) [160 - 2W(2 + pi)]. Multiply both sides by 8,
0 = 160 - 2W(2 + pi). Now, solve for W.

2W(2 + pi) = 160
W = 160 / [2(2 + pi)]
W = 80 / (2 + pi)

This is WHERE the maximum area occurs. To actually find WHAT the maximum area is, we plug in A(80 / (2 + pi)) into our function.

A(W) = (1/8) [160W - (4 + pi)W^2]
A(80 / (2 + pi)) = (1/8) [160 {80 / (2 + pi)} - (4 + pi) [80 / (2 + pi)]^2]

I'll leave it to you to solve the rest.

The general methology for optimization:

(1) Find an equation where you can express one variable in terms of the other,
(2) Create your maximizing function that will depend on ONE variable.

Answer 4

Area of shape = area of rectangle + area of semi-circle

semi-circle = half of a cirle

So,
A = (l * w) + ((pi*r*r)/2)

Note: The width of the rectangle is the same as the diameter (or 2r) of the circle.

Then,
A = (l * 2r) + ((pi*r*r)/2)

We have 3 variables here, and I'm guessing you can't solve that yet. So we have to find another equation that relates two of the variables.

We've been given a hint. 40 m of fencing. So let's look at the perimeter of the shape.

First I'd look at the semi-circle part of the shape.

If C = 2(pi)r then half of a circle would have a perimeter of (pi)r.

Then you have 3 remaining sides: l, l, and 2r

Since you have 40 m of fence then:

40 = l + l + 2r + (pi)r

Let's solve this equation for l:

2l = 40 - 2r - (pi)r
l = 20 -r -(pi/2)r

Let's plug this into our area equation A = (l * 2r) + ((pi*r*r)/2):

A = ((20 -r -(pi/2)r)* 2r) + ((pi*r*r)/2)

From here, you need to find the maximum (or y coordinate of the vertex)
-you can use a TI-83 series calculator and the max function
-a table of values
-graph it by hand

Hope that helps.

Answer 5

put everything in terms of each other (radius to lenghts, etc). Then differentiate the equation of the total area and make it equal to 0.