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.

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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?

Your solution:

Confidence Assessment:

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. **

STUDENT QUESTION:

I got the formula for the area but don't understand how you get a parabola out of that.

INSTRUCTOR RESPONSE:

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.

Self-critique (if necessary):

Self-critique Rating:

Question: *  Extra Problem / 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, a third on the positive x-axis, and a fourth vertex on the positive y-axis. Find the largest area that can be enclosed by the rectangle.

Your solution:

Confidence Assessment:

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. **

Self-critique (if necessary):

Self-critique Rating: