Optimization Problem

1

Optimization Problem_

Collecting Data:

We want to maximize the amount of light that can come through the window, so what we are looking for is the area of the window. We want to know what dimensions of the rectangle will give us the largest area with the perimeter of the rectangular part being 12 feet. With collecting data I would like you to start with your independent variable and plug in different numbers for the window. Construct a table of the x and y values of the rectangle and determine the area for each x and y value that you choose.

Solution:

The first step in this problem is to find/draw what the window will look like:

Now it is important to establish the relationship between the x, y, and d values. We know that the perimeter of the rectangle is equal to 12 feet and the diameter of the circle is equal to the diameter:

2x+2y=12 ft.

d=y

Since the diameter of the circle is equal to the length (y) of the rectangle, that becomes the dependent variable. Now we can start plugging in different values for x and y:

2x+2y=12 [divide each side by 2]

x+y=6 [Now we can plug in the different values for x and y and create a table]

Since we know that neither x or y can be equal to 0, we can eliminate the sets (6,0) and (0,6). Let’s start making our table of integers as well as determining the area for each case:

X / Y / Area
1 / 5 / 5
2 / 4 / 8
3 / 3 / 9
4 / 2 / 8
5 / 1 / 5

Now we can use this information to calculate the area of the semi-circle:

Area of a circle = [remember r = ½ *d]

Since we know the values of d are equal to y, we can make another table including the radius of the circle and the area of the circle, but before doing so make a guess as to which values of x and y you think will generate the largest total area:

[here the students would take a guess as to which they think the largest area would be]

Now make the table:

X / Y / Area of rectangle / R / Area of Circle
1 / 5 / 5 / 2.5 / 6.25π
2 / 4 / 8 / 2 / 4π
3 / 3 / 9 / 1.5 / 2.25π
4 / 2 / 8 / 1 / π
5 / 1 / 5 / .5 / .25π

Now remember, we only want the area of half of the circle so multiply each of the areas by ½ . From this information we can now determine for which of these x and y values will produce the most amount of sunlight by adding the areas together:

X / Y / Area of rectangle / R / Area of Circle / Sum of Areas
1 / 5 / 5 / 2.5 / 3.125π / 14.82
2 / 4 / 8 / 2 / 2π / 14.28
3 / 3 / 9 / 1.5 / 1.125π / 12.53
4 / 2 / 8 / 1 / 0.5π / 9.57
5 / 1 / 5 / .5 / 0.125π / 5.39

We can clearly see that when x is 1 and y is 5, the area of the window will be the greatest, meaning the most amount of light will pass through.

Function modeling the data:

Using the information from the problem of the Norman window, find a function that would model the data in order to determine the x and the y values that will provide the largest amount of light passing through.

Solution:

We need to start by looking at what information that we have. We know that the perimeter of the rectangle is equal to 12 ft. and the diameter of the semicircle is equal to the width of the rectangle:

Since we know this we can start by formulating an equation that will determine the area of the rectangle starting with the equation for the perimeter:

2x + 2y = 12[divide both sides by 2]

x + y = 6[solve for y]

y = -x + 6

Now we need to find the equation for the area:

x * y = A[set area = A]

Now we need the area of a circle:

Area of a circle = πr²

Since we know that r = ½d, and that d = y, we can substitute into the equation:

π(1/2 * y)²

Sicne we only want to know the area of half of the circle, we need to multiply both sides by ½ :

½ π(1/2*y)²

We want to know the total area of the window so now we can add the two areas together:

x*y + ½ π(1/2*y)² = Area of the window

Now we can substitute in for y and round π to the nearest hundreth:

x(-x + 6) + ½ π (1/2(-x + 6))²[use the distributive property]

-x² + 6x + ½ π(-x/2 + 3)²[use the foil method]

-x² + 6x + ½ π(1/4x² - 3x + 9)[round π and use the distributive property again]

-x²+ 0.39x² + 6x – 4.71x + 14.14[now use algebra to clean up this function]

-0.61x² + 1.29x + 14.14

Now we have a function that we can plug into the calculator. Set the Y₁ equal to the function and look at the graph:

Now we know that neither x or y can equal 0, so we can ignore any parts of the graph that cross into the negatives of the x and y-axis:

What is it that you notice about the graph, does it support the answer from the first question using data collection? In fact it does, the graph peaks very early on and as the x-values get larger, the y values get smaller and the area decreases (which is what we decided in the first part of the problem). Now you can use the calculator to find the maximum value of the curve. This will tell you which x and y values will give you the largest area.

Find the maximum by pressing 2nd,CALC, and then 4 for maximum. Once you do this the calculator will ask you to place to points on the curve, left bound and right bound. Place the left bound point left of where you believe the max to be (don’t forget to place it a little far away in order to account for any errors). Do the same with the right bound, except place it to the right of where the max is, and then it will ask you to guess. Trace up the finction until you get close to where the max is and then press ENTER. The calculator will then calculate the maximum based on the information that you gave it:

Maximum is at the point (1.06,14.82).

So we now know the maximum of the graph is at the point x = 1.06 and the area is equal to 14.82. Does this support your answer from the first part of the question? Indeed it does. Don’t forget that we rounded both times so the answer is going to be a little bit off, but still close.

TI-Nspire:

Now we will use the TI-Nspire to find the x and y values in order to maximize the amount of light that passes through the window.

Solution:

What we want to do is plug the function into the calculator and view the graph of the function. First go to the home page of the calulator:

Now either highlight New Documents and press ENTER, or press 5:

Now let’s select the Add Graphs & Geometry function by highlighting and pressing ENTER, or by pressing 2:

Now the cursor should already be next to the function box (f1(x)=). Now let’s type in the function that we previously found and press enter:

As you can see, the graph is cut off so we need to adjust our window. While we adjust the window, we should also remember that we don’t need any of the values that cross the x or y-axis because x>0 and y>0. So let’s adjust the window. Start by pressing MENU:

Select option 4: Window:

Now select the first option:

Let’s change the XMin to 0, the XMax to 10, the YMin to 0, and the YMax to 20, and then highlight OK:

Press enter once you have made these changes:

Now we have a clear view of our graph. We want to find the maximum of this graph, this will tell us at what point the area of our window will be maximized. Start by hitting the MENU button again and then selecting 6:Points & Lines:

Select option 2:Point On:

Now move the arrow over to a point on the curve until you see what looks like a pencil:

Press ENTER:

Now press enter until the arrow appears on your screen again:

Now move the arrow over to the point until the hand appears and press and hold the action button (the button with the hand on it) until the hand looks like it is grabbing the point:

Now move the point toward where the max of the graph is. Once you hit the max, a little M in a box will appear. This is showing that you have found the maximum value of the graph:

Now press ENTER and record the x and y values that you see:

We see that x = 1.06 and y = 14.22. What is it about these values that you notice? They are consistent with what we have found in the past 2 parts of this problem. Therefore, the maximum amount of light will pass through when x = 1.06 and the area = 14.22.

Calculus:

Using the information from the problem of the Norman window, use calculus to find the maximum area of the window. We want to find the x and y values that will allow the most light to pass through the window.

Solution:

We need to first look at the information we have. We know the perimeter of the rectangle is equal to 12 feet and the diameter of the semicircle is equal to the width of the rectangle:

(Insert your diagram)

Since we know this we can start by formulating an equation that will determine the area of the window. We want to know the total area of the window, so we can use our previous answer to complete the next step. We know:

x*y + ½ (½ * y)^2 = Area of window [from your previous work]

Now we need to substitute for y:

X(-x + 6) + ½ (½ (-x + 6))^2 [use the distributive property]

-x^2 + 6x + ½ (-x/2 + 3)^2 [use the foil method]

-x^2 + 6x + ½  (¼ x^2 - 3x + 9) [use the distributive property again]

-x^2 + 6x +/8 x^2 - 3/2 x + 9/2 [now use algebra to clean up the function]

-7/8 x^2 + 9/2 x + 9/2

Now we have a function that we can use to find the maximum area of the window. First, we need to find the derivative of the function:

A = -7/8 x^2 + 9/2 x + 9/2 [now find the derivative of this equation]

A = -7/8 (2)x + 9/2 [now simplify the expression using algebra]

A = -7/4 x + 9/2

Now we have a function that we can solve to find the maximum area of the window. We need to set the equation equal to zero and solve for x:

-7/4 x + 9/2 = 0[subtract 9/2 from each side]

-7/4 x = - 9/2 [divide both sides by -7/4]

X = 18/7

We have found the x-value to be 2.57. Now to find the y-value we need to put 2.57 in for x in the equation y = -x +6:

Y = -x + 6[put 2.57 in for x]

Y = -2.57 + 6[solve]

Y = 3.43

We now know both the x and y values. We need to put these values in to our area formula to find the maximum amount of light:

A = x*y + ½ (½ * y)^2 [put in the x and y]

A = 2.57 * 3.43 + /2(½ * 3.43)^2[solve]

A = 13.44 square feet

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