course mth 158
If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
Your solution, attempt at solution:
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
031. * 31
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Question: * 4.4.10 / 7th edition 4.1.78 (was 4.1.60). A farmer with 2000 meters of fencing wants to enclose a rectangular plot that borders on a straight highway. If the farmer does not fence the side along the highway, what is the largest area that can be enclosed?
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Your solution:
length = x
fence = 2000 – x
width = 2000 – x / 2
Area = lw
A = x (2000 – x / 2)
A = 2000x – x^2 / 2
A = 1000x – (x^2/2)
- 1/2x^2 + 1000x now we have a quadratic equation and can solve for x
max = -b/2a = - 1000 / 2(-1/2) = 1000
This is as far as I got with my tutor. I don’t think he understood it that well. I really don’t think I could solve this again on my own.
confidence rating: 0
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Given Solution:
* * If the farmer uses 2000 meters of fence, and fences x feet along the highway, then he have 2000 - x meters for the other two sides. So the dimensions of the rectangle are x meters by (2000 - x) / 2 meters.
The area is therefore x * (2000 - x) / 2 = -x^2 / 2 + 1000 x.
The graph of this function forms a downward-opening parabola with vertex at x = -b / (2 a) = -1000 / (2 * -1/2) = 1000.
At x = 1000 the area is -x^2 / 2 + 1000 x = -1000^2 / 2 + 1000 * 1000 = 500,000, meaning 500,000 square meters.
Since this is the 'highest' point of the area vs. dimension parabola, this is the maximum possible area. **
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Self-critique (if necessary):
I have went over and over this and just cannot comprehend how to do these problems.
There are two overall steps to solving this problem. The first is to get an expression for the area. The second is to find the maximum possible value of the expression.
You have the first step, which gives you the equation A = 1000 x - x^2 / 2.
The second step is to find the maximum of your expression.
To find the maximum of the expression 1000 x - x^2 / 2, you can consider the graph of the function.
The expression can be written as - x^2 / 2 + 1000 x. This is in the form of a quadratic expression of form
a x^2 + b x + c with
a = -1/2, b = 1000 and c = 0.
As you know from your previous work a quadratic function has a graph which is a parabola, and the vertex of a parabola is either its highest or lowest point (depending on whether it opens downward or upward).
The vertex of the parabola occurs when x = - b / (2 a). Substituting b = 1000 and a = -1/2 we find that
x_vertex = - 1000 / (2 * (-1/2) ) = 1000. The corresponding y coordinate is
y = -1/2 x_vertex^2 + 1000 * x_vertex = -1/2 * 1000^2 + 1000 * 1000 = 500 000.
Since a = -1/2, which is negative, the parabola opens downward. This makes the vertex the highest point, so the value of the function is maximized at the vertex.
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Self-critique Rating: 0
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Question: * 4.4.33 / 7th edition 4.1.101 (was 4.1.80). A rectangle has one vertex on the line y=10-x, x>0, another at the origin, one on the positive x-axis, and one on the positive y-axis. Find the largest area that can be enclosed by the rectangle.
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Your solution:
This is another one that my tutor and I discussed. He started with a graph that had 10 units on each axis, then drew a line from y = 10 to x = 10 to form a triangle on the graph. Then he drew a rectangle inside the triangle with the length = x and the width = 10 – x to solve for
A = lw
= x(10-x)
= -x^2 + 10x (all of this made perfect sense when I saw him do it.)
then to find the maximum
-b/2a = -10/2(-1) = -10/-2 = 5
then square that to get 5 * 5 = 25 square units
(He also drew a table of x vs. y with the x starting at 0 going to 10 and the y starting at 10 and going to 0. This is how he explained the y = 10 – x. After I look at it, I can see how 5 is the only point on the table where x = y)
confidence rating: 3
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Given Solution:
* * The dimensions of the rectangle are x and y = 10 - x. So the area is
area = x ( 10 - x) = -x^2 + 10 x.
The vertex of this rectangle is at x = -b / (2 a) = -10 / (2 * -1) = 5.
Since the parabola opens downward this value of x results in a maximum area, which is
-x^2 + 10 x = -5^2 + 10 * 5 = 25. **
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Self-critique (if necessary):
There are no discrepancies.
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Self-critique Rating: Without the tutor’s help, I don’t think I would have been able to solve this one.
&#Good responses. See my notes and let me know if you have questions. &#