Chase Quinn wants to expand his cut-flower business. He has 12 additional acres on which he intends to plant lilies and gladioli. He can plant at most 7 acres of gladiolus bulbs and no more than 11 acres of lilies. In addition, the number of acres planted to gladioli G can be no more than twice the number of acres planted to lilies L. The inequality L + 2G > 10 represents his labor restrictions. If his profits are represented by the function f(L, G) = 300L + 200G, how many acres of lilies should he plant to maximize his profit?
To maximize his profit, Chase Quinn needs to find the optimal number of acres of lilies to plant. We can do this by using the constraints provided and applying the concept of linear programming.
Let's break down the problem:
1. Constraints:
- He can plant at most 7 acres of gladioli: G <= 7
- He can plant no more than 11 acres of lilies: L <= 11
- The number of acres planted to gladioli can be no more than twice the number of acres planted to lilies: G <= 2L
- Labor restrictions: L + 2G > 10
2. Objective function:
- Chase's profits are represented by the function f(L, G) = 300L + 200G, where L represents the number of acres planted to lilies and G represents the number of acres planted to gladioli.
Now, we need to determine the feasible region, which is the overlapping area of all constraints.
To simplify the calculations, let's convert the inequalities into equalities:
1. G = 7 (since he can plant at most 7 acres of gladioli)
2. L = 11 (since he can plant no more than 11 acres of lilies)
3. G = 2L (since the number of acres planted to gladioli can be no more than twice the number of acres planted to lilies)
Now that we have these three equations, we can solve them by substitution:
1. G = 7
2. L = 11
3. 7 = 2L
Substituting equation 3 into equation 1:
7 = 2(11)
7 = 22
Since 7 = 22 is not true, there is no common solution that satisfies all three constraints. Therefore, the feasible region does not exist.
However, to find the maximum number of acres of lilies for profit maximization, we can use the labor restriction constraint: L + 2G > 10.
Substituting the values for G:
L + 2(7) > 10
L + 14 > 10
L > 10 - 14
L > -4
Therefore, the number of acres of lilies that Chase should plant to maximize his profit is greater than -4. However, since lilies cannot be planted in negative acreage, the maximum number of acres of lilies he can plant is 0.
To maximize Chase Quinn's profit, we need to find the values for L (acres of lilies) and G (acres of gladioli) that satisfy all of his restrictions and yield the highest profit according to the function f(L, G) = 300L + 200G.
First, let's list the restrictions:
1. Chase has 12 additional acres available for planting.
2. He can plant at most 7 acres of gladiolus bulbs.
3. He can plant no more than 11 acres of lilies.
4. The number of acres planted to gladioli (G) can be no more than twice the number of acres planted to lilies (L).
5. The inequality L + 2G > 10 represents his labor restrictions.
To determine the maximum profit, we will create a feasible region by graphing these restrictions on a coordinate plane.
- Chase has 12 additional acres, which means L + G ≤ 12.
- The maximum limit for gladiolus is 7 acres, so G ≤ 7.
- The maximum limit for lilies is 11 acres, so L ≤ 11.
- The inequality L + 2G > 10 can be simplified to G < (L + 10) / 2.
Now, we will graph these inequalities and identify the feasible region.
1. Plotting the lines L + G = 12, G = 7, L = 11:
- Draw the line L + G = 12 (a diagonal line from (0, 12) to (12, 0)).
- Draw a horizontal line at G = 7.
- Draw a vertical line at L = 11.
2. Shading the feasible region:
- Shade the region below the line L + G = 12 (including the line).
- Shade the region below the horizontal line at G = 7 (including the line).
- Shade the region to the left of the vertical line at L = 11 (including the line).
- Shade the region below the line G < (L + 10) / 2.
The shaded region represents the feasible region, satisfying all of the given restrictions.
Next, we need to find the points of intersection within the feasible region.
- The intersection points of L + G = 12 and G = 7 are (5, 7) and (7, 5).
- The intersection point of L + G = 12 and L = 11 is (11, 1).
- The intersection point of L + G = 12 and G = (L + 10) / 2 is (4, 8).
Next, we substitute the coordinates from the intersection points into the profit function f(L, G) = 300L + 200G to determine the profit for each point.
- f(5, 7): 300(5) + 200(7) = 2500.
- f(7, 5): 300(7) + 200(5) = 2900.
- f(11, 1): 300(11) + 200(1) = 3400.
- f(4, 8): 300(4) + 200(8) = 3200.
Comparing the profits, we can see that the highest profit is obtained when Chase plants 11 acres of lilies and 1 acre of gladioli. Therefore, he should plant 11 acres of lilies to maximize his profit.