Mr. Green, the Herbarium's owner, has come to you with another problem to solve. He wants to create a new sales display featuring giant potted plants and plants in smaller hanging baskets. He wants to set up the display in such a way that he can maximize his profit from that area, but doesn't know how many of each plant type to use. He provides you with certain parameters that limit the number of potted plants and hanging baskets he can use in the display. Use the following information to tell Mr. Green how many giant potted plants and how many plants in hanging baskets he should display to maximize his profits.
Mr. Green wants to dedicate no more than 168 square feet to the display. Each giant potted plant requires 7 square feet of space. Each hanging basket requires 4 square feet of space. The Herbarium has 210 gallons of soil available to pot the plants for the display. Giant potted plants need 14 gallons of soil each while the hanging baskets need 2 gallons of soil each. Mr. Green has allocated $252 to buy the plants for this display. Giant potted plants cost $14 each while the plants for hanging baskets cost $5 each. The Herbarium makes a profit of $28 on each giant potted plant sold, and a profit of $8 on each plant in a hanging basket sold.
A. Declare a set of variables pertaining to Mr. Green's situation and write a system of inequalities that models Mr. Green's situation. There are directions for how to type math characters in the Resource Center that will show you show to enter your inequalities
--p= giant potted plants h= hanging baskets
x= p y= h
7p + 4h <= 168 sq. ft.
7x + 4y <= 168
14p + 2h <= 210gallons of available soil.
14x + 2y <= 210
$14p + $5h <= $252
$14x + $5y <= $252
P= 28p + 8h
B. Describe the solution graph by giving the boundaries of the solution area as inequalities and state the relationship of the solution to the inequality, i.e.: above, below, to the left, to the right, etc.
-- ?!?!?!? help?
Part A:
The inequalities of part A are correct.
Missed out are the conditions
x≥0 and y≥0.
The objective function P=28P+8H is doubtful, because there is no statement saying that sales are proportional to the number of samples on display.
Part B:
You will have to plot the graphs and find out what the lines look like. Since all your inequalities are ≤, so the feasible regions are all below the lines plotted. In fact, the final feasible region will be the polygon below all the lines plotted, and above the x-axis, and to the right of the y-axis.
You will find four vertices at the top of the feasible region, places where solutions are most likely to lie.
You will evaluate the objective function at these four vertices and choose the most favourable one.
As a check, these vertices are associated with values of x of 0,8,13 and 15.
Use this paragraph to answer the question.
Marta was nervous about taking care of her neighbor's potted plants. She had no experience with gardening, and her family never kept plants inside the house due to her mother's allergies. Now, in Mrs. Brown's sunroom that was crowded with more than 20 live and exotic plants, she began to regret taking the job. Marta looked over the notes that were left for her on the potting table and rubbed her forehead. She had to read the first page three times before she understood how to feed the orchids.
Which piece of textual evidence supports the inference that Mrs. Brown's directions for Marta were complicated?
(1 point)
Responses
Marta was nervous.
Marta was nervous.
Marta had to read the first page three times.
Marta had to read the first page three times.
She understood how to feed the orchids.
She understood how to feed the orchids.
Mrs. Brown's sunroom was crowded.
To describe the solution graph, we need to find the boundary lines for the solution area.
1. For the space constraint, we have the inequality: 7p + 4h <= 168 sq. ft.
- The boundary line for this inequality is: 7p + 4h = 168.
2. For the soil constraint, we have the inequality: 14p + 2h <= 210 gallons of available soil.
- The boundary line for this inequality is: 14p + 2h = 210.
3. For the budget constraint, we have the inequality: $14p + $5h <= $252.
- The boundary line for this inequality is: $14p + $5h = $252.
Now, let's plot these boundary lines on a graph:
The first boundary line, 7p + 4h = 168, represents the space constraint. To plot it, we need to find two points that satisfy this equation.
- When p = 0, we have 4h = 168 => h = 42. So, one point is (0, 42).
- When h = 0, we have 7p = 168 => p = 24. So, another point is (24, 0).
Plotting these two points and drawing a line through them will represent the space constraint.
The second boundary line, 14p + 2h = 210, represents the soil constraint. Similarly, we find two points that satisfy this equation.
- When p = 0, we have 2h = 210 => h = 105. So, one point is (0, 105).
- When h = 0, we have 14p = 210 => p = 15. So, another point is (15, 0).
Plotting these two points and drawing a line through them will represent the soil constraint.
The third boundary line, $14p + $5h = $252, represents the budget constraint. We also find two points that satisfy this equation.
- When p = 0, we have $5h = $252 => h = 50.4 (let's round it to 50). So, one point is (0, 50).
- When h = 0, we have $14p = $252 => p = 18. So, another point is (18, 0).
Plotting these two points and drawing a line through them will represent the budget constraint.
The solution area will be the region that satisfies all three inequalities, which is the area below all three lines (since the inequalities are less than or equal to).
To summarize, the boundaries of the solution area in the graph are:
- The line 7p + 4h = 168 (representing the space constraint)
- The line 14p + 2h = 210 (representing the soil constraint)
- The line $14p + $5h = $252 (representing the budget constraint)
The solution area will be below all three lines.
To describe the solution graph, we need to find the boundaries of the solution area.
Let's start by graphing the first inequality: 7p + 4h <= 168.
To graph this inequality, we need to rewrite it in slope-intercept form: h <= (-7/4)p + 42.
The boundary line for this inequality has a slope of -7/4 and a y-intercept of 42.
Next, let's graph the second inequality: 14p + 2h <= 210.
We can rewrite this inequality as: h <= (-14/2)p + 105.
The boundary line for this inequality has a slope of -14/2 and a y-intercept of 105.
Finally, let's graph the third inequality: $14p + $5h <= $252.
We can rewrite this inequality as: h <= (-14/5)p + 50.4.
The boundary line for this inequality has a slope of -14/5 and a y-intercept of 50.4.
Now, we can plot the boundaries of the solution area on a graph and determine the relationship of the solution to the inequalities. The solution area will be below the three boundary lines.
I hope this helps!