A factory can produce two products, x and y, with a profit approximated by P = 14x + 22y – 900. The production of y can exceed x by no more than 200 units. Moreover, production levels are limited by the formula x + 2y ≤ 1600. What production levels yield maximum profit? (1 point) Responses x = 400 y = 600 x = 400 y = 600 x = 0 y = 0 x = 0 y = 0 x = 1600 y = 0 x = 1600 y = 0 x = 0 y = 200
To find the production levels that yield maximum profit, we need to find the values of x and y that maximize the profit function P = 14x + 22y - 900, subject to the given constraints.
The first constraint states that the production of y can exceed x by no more than 200 units. We can write this constraint mathematically as:
y - x ≤ 200
The second constraint states that the total production levels are limited by the formula x + 2y ≤ 1600.
Now, we can solve this problem using the method of linear programming:
Step 1: Write the objective function:
P = 14x + 22y - 900
Step 2: Write the constraints:
y - x ≤ 200
x + 2y ≤ 1600
Step 3: Convert the inequalities to equations:
y - x = 200
x + 2y = 1600
Step 4: Solve the system of equations:
Using the method of substitution, we can solve for x in terms of y:
From the first equation, we rearrange it as:
x = y - 200
Substituting this value of x into the second equation, we get:
y - 200 + 2y = 1600
3y - 200 = 1600
3y = 1800
y = 600
Substituting the value of y back into the first equation, we get:
x = 600 - 200
x = 400
Therefore, the values that yield maximum profit are:
x = 400 and y = 600.
So, the correct response is:
x = 400
y = 600
To find the production levels that yield the maximum profit, we need to solve the given system of inequalities.
x + 2y ≤ 1600 (restricting the production levels)
y - x ≤ 200 (limiting the difference between y and x)
We can solve this system by graphing or using linear programming techniques. Let's use graphing in this case.
First, let's graph the first inequality x + 2y ≤ 1600:
Plotting the x and y-intercepts:
To find the x-intercept, set y = 0:
x + 2(0) = 1600
x = 1600
So, the x-intercept is (1600, 0).
To find the y-intercept, set x = 0:
0 + 2y = 1600
2y = 1600
y = 800
So, the y-intercept is (0, 800).
Plotting these two points, we can draw a line that passes through them.
Next, let's graph the second inequality y - x ≤ 200:
Rewrite the inequality as y ≤ x + 200.
Plotting the x and y-intercepts:
To find the x-intercept, set y = 0:
0 ≤ x + 200
x ≥ -200
So, the x-intercept is (-200, 0).
To find the y-intercept, set x = 0:
y ≤ 0 + 200
y ≤ 200
So, the y-intercept is (0, 200).
Plotting these two points, we can draw a line that passes through them.
Now, we shade the region that satisfies both inequalities, which is the overlapping region between the two lines.
In this shaded region, we need to find the production levels that maximize the profit function P = 14x + 22y - 900.
To do that, we can evaluate the profit function at the vertices of the shaded region.
The vertices are (0, 0), (1600, 0), and (400, 600).
Evaluating the profit function at these vertices:
P(0, 0) = 14(0) + 22(0) - 900 = -900
P(1600, 0) = 14(1600) + 22(0) - 900 = 22,400 - 900 = 21,500
P(400, 600) = 14(400) + 22(600) - 900 = 5,600 + 13,200 - 900 = 18,900
The maximum profit is achieved at (400, 600) with a profit of 18,900.
So, the production levels that yield maximum profit are x = 400 and y = 600.
To determine the production levels that yield maximum profit, we need to optimize the profit function while considering the given constraints.
Let's break down the problem step by step:
1. Define the variables:
Let x represent the production level of product x (in units), and y represent the production level of product y (in units).
2. Formulate the objective function:
The profit function is given by P = 14x + 22y - 900. Our goal is to maximize this function.
3. Incorporate the constraints:
a) The production level of y can exceed x by no more than 200 units: y ≤ x + 200.
b) The production levels are limited by x + 2y ≤ 1600.
4. Set up the linear programming problem in standard form:
Maximize P = 14x + 22y - 900
Subject to:
- y ≤ x + 200
- x + 2y ≤ 1600
- x ≥ 0
- y ≥ 0
5. Solve the linear programming problem:
Using a linear programming solver or graphing the feasible region and evaluating the objective function at each corner point, we can find the optimal solution.
The optimal solution to this linear programming problem is:
x = 400
y = 600
These production levels yield the maximum profit.