A breed of cattle needs at least 10 proteins and 8 fats units per day. Feed type I provides 6 protein and 2 fat units at $4 per
1 answer:
Let x represent the amount of feed type I and y the amount of feed type II to be utilized. Our goal is to minimize the following:
C = 4x + 3y
while adhering to these constraints:
Analyzing the graph of these four constraints yields corner points at (0, 5), (1, 2), and (4, 0).
Testing the objective function at each corner point to find the minimum:
For (0, 5):
C = 4(0) + 3(5) = $15
For (1, 2):
C = 4(1) + 3(2) = 4 + 6 = $10
For (4, 0):
C = 4(4) + 3(0) = $16.
Thus, the combination that results in the lowest cost is one bag of feed type I and two bags of feed type II.
C = 4x + 3y
while adhering to these constraints:
Analyzing the graph of these four constraints yields corner points at (0, 5), (1, 2), and (4, 0).
Testing the objective function at each corner point to find the minimum:
For (0, 5):
C = 4(0) + 3(5) = $15
For (1, 2):
C = 4(1) + 3(2) = 4 + 6 = $10
For (4, 0):
C = 4(4) + 3(0) = $16.
Thus, the combination that results in the lowest cost is one bag of feed type I and two bags of feed type II.
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Answer:
The amount of ovens that must be produced in a week to earn a $1610 profit is 70.
Step-by-step explanation:
Given:
A small-appliance manufacturer determines the profit P (in dollars) from producing x microwave ovens weekly by the formula:
with 0 ≤ x ≤ 200.
The target profit is $1610
So, set P = 1610, then solve for x:
Multiply both sides by 10:
Next, factor the quadratic:
Solving for x gives:
Since x=230 is outside the domain 0 ≤ x ≤ 200, we discard it.
Hence, the valid solution is x=70.
Therefore, to achieve a $1610 profit, the manufacturer must produce 70 ovens weekly.