A trucker hauls citrus fruit from Florida to Montreal. Each crate of oranges is 4 ft3 in volume and weighs 60 lb. Each crate of grapefruit has a volume of 6 ft3 and weighs 75 lb. His truck has a maximum capacity of 300 ft3 and can carry no more than 4200 lb. Moreover, he is not permitted to carry more crates of grapefruit than crates of oranges. If his profit is $2.50 on each crate of oranges and $4 on each crate of grapefruit, how many crates of each fruit should he carry for maximum profit?
Incorrect: Your answer is incorrect.
oranges
Incorrect: Your answer is incorrect.
grapefruits
oh, if you graphed carefully, no need to do (31.11,31.11) because it is above others. however the (70,0) point where b hits the x axis should be checked even if you can see it is not going to be max.
Well, it seems like the trucker is in a bit of a pickle, or should I say orange? To maximize his profit, he needs to figure out the perfect combination of oranges and grapefruits to carry in his truck.
Let's do some math-fun to solve this problem. Let's say the number of crates of oranges is 'o' and the number of crates of grapefruits is 'g.'
We know that each crate of oranges weighs 60 lb and has a volume of 4 ft3, while each crate of grapefruits weighs 75 lb and has a volume of 6 ft3.
To stay within the truck's maximum capacity, we have the following constraints:
4o + 6g ≤ 300 (volume constraint)
60o + 75g ≤ 4200 (weight constraint)
g ≤ o (the number of crates of grapefruits should be less than or equal to the number of crates of oranges)
Now, let's have some clown-fun and find the optimal solution!
After some calculations and clown-trickery, we find that the best combination to maximize profit is:
oranges: 30 crates
grapefruits: 20 crates
So, my dear trucker friend, load up the truck with 30 crates of oranges and 20 crates of grapefruits, and watch the profit roll in like a citrusy avalanche!
To maximize profit, the trucker should carry as many crates of grapefruit as possible while staying within the truck's capacity constraints.
Let's assume the trucker carries x crates of oranges and y crates of grapefruit.
The volume constraint can be represented by the equation:
4x + 6y ≤ 300 (in ft^3)
The weight constraint can be represented by the equation:
60x + 75y ≤ 4200 (in lb)
Additionally, we know that the number of crates of grapefruit cannot exceed the number of crates of oranges, so y ≤ x.
To solve this problem, we can consider the profit function:
Profit = (2.50 * x) + (4 * y)
We want to find the values of x and y that maximize the profit function while satisfying the constraints.
Now, let's solve the problem step-by-step.
Step 1: Rewrite the volume constraint in terms of one variable:
4x + 6y ≤ 300
Divide all terms by 2 to simplify:
2x + 3y ≤ 150
Step 2: Rewrite the weight constraint in terms of one variable:
60x + 75y ≤ 4200
Divide all terms by 15 to simplify:
4x + 5y ≤ 280
Step 3: Combine the volume constraint (2x + 3y ≤ 150) and the weight constraint (4x + 5y ≤ 280) for graphical representation. Plot the relevant line segments and find the feasible region.
Step 4: Determine the vertices of the feasible region.
By solving the system of linear equations:
2x + 3y = 150
4x + 5y = 280
We find:
x = 50, y = 0 (Point A)
x = 0, y = 56 (Point B)
x = 20, y = 40 (Point C)
Step 5: Evaluate the profit function at each vertex:
At Point A: Profit = (2.50 * 50) + (4 * 0) = $125
At Point B: Profit = (2.50 * 0) + (4 * 56) = $224
At Point C: Profit = (2.50 * 20) + (4 * 40) = $200
Step 6: Compare the profits and determine the maximum profit:
The maximum profit is $224, achieved at Point B.
Therefore, the trucker should carry 0 crates of oranges and 56 crates of grapefruit for maximum profit.
Correct: The trucker should carry 56 crates of grapefruit for maximum profit.
Incorrect: The trucker should not carry any crates of oranges for maximum profit.
To find the number of crates of each fruit that the trucker should carry for maximum profit, we can use a mathematical technique called linear programming.
Let's assume the trucker carries x crates of oranges and y crates of grapefruits.
The volume constraint can be represented as:
4x + 6y ≤ 300 (since the truck's maximum capacity is 300 ft^3)
The weight constraint can be represented as:
60x + 75y ≤ 4200 (since the truck's maximum weight capacity is 4200 lbs)
Since the trucker is not permitted to carry more crates of grapefruit than crates of oranges, we also have the constraint:
y ≤ x
The profit function can be defined as:
Profit = 2.50x + 4y (per crate profit for oranges and grapefruits respectively)
To find the maximum profit, we need to maximize the profit function subject to the given constraints.
One method to solve this problem is by graphing the feasible region (the region that satisfies all the constraints) and finding the corner point with the highest profit.
However, since finding the feasible region and graphing can be time-consuming, we can use a simpler method called the "Corner Point Method."
By calculating the values of the profit function at the corner points of the feasible region, we can determine which combination of crates gives the maximum profit.
The corner points of the feasible region can be found by solving the system of linear equations for the inequalities.
1) Let's solve the volume constraint for x in terms of y:
4x + 6y ≤ 300
4x ≤ 300 - 6y
x ≤ 75 - (3/2)y
2) Let's solve the weight constraint for x in terms of y:
60x + 75y ≤ 4200
60x ≤ 4200 - 75y
x ≤ 70 - (5/4)y
3) Now, let's graph the region satisfying the constraint y ≤ x as a line: y = x.
4) Now, we can calculate the profit function, Profit = 2.50x + 4y, at the corner points of the feasible region determined by the intersection of lines and the inequality constraints.
By evaluating the profit function at each corner point, we can determine which combination of crates of oranges and grapefruits yields the maximum profit.
Since you mentioned that the answer provided is incorrect, there might be an error in the calculations or assumptions made.
orange crates = x
grapefruit crates = y
constraints
a)4x+6y</= 300
b)60 x + 75 y</= 4200
c)y</= x
maximize p = 2.5 x + 4 y
graph those constraint lines and mark intersections
a hits c at (30,30)
b hits c at (31.11, 31.11)
a hits b at (45,20)
our region is between the x axis and those lines, check intersections for max p
p = 2.5 x + 4 y
(30,30) --> p = 195
(31.11,31.11) --> p = 202.2
(45,20) -> p = 192.5
(70,0) -> P = 175
SO 45 ORANGE, 20 GRAPEFRUIT