QUESTION 4
[2+2+3+3 = 10 marks]
A dietician has two foods with which to prepare a diet containing at least 400 units of calcium, 175 units of iron, and 280 units of vitamin A. The table below gives the units of calcium, iron, and vitamin A per ounce for the two foods.
Food I costs 35 cents an ounce and Food II costs 42 cents an ounce. What combination of the two foods gives the diet, meeting the above requirements, of minimum costs?
no table.
If you care to append it now, we can offer some help, but I suspect this problem is very like others you have already done, so ...
To determine the combination of foods that meets the requirements at minimum cost, we need to set up an optimization problem.
Let's assign variables to the quantities of Food I and Food II used in the diet. Let:
- x be the number of ounces of Food I
- y be the number of ounces of Food II
We can set up equations based on the given requirements:
The calcium requirement: x units of calcium per ounce * x ounces of Food I + y units of calcium per ounce * y ounces of Food II ≥ 400 units of calcium
=> x + y ≥ 400 / x + 420 / y ≥ 400 (equation 1)
The iron requirement: x units of iron per ounce * x ounces of Food I + y units of iron per ounce * y ounces of Food II ≥ 175 units of iron
=> x + 3y ≥ 175 (equation 2)
The vitamin A requirement: 2x units of vitamin A per ounce * x ounces of Food I + 3y units of vitamin A per ounce * y ounces of Food II ≥ 280 units of vitamin A
=> 2x + 3y ≥ 280 (equation 3)
To minimize the cost, we need an objective function. The total cost can be calculated as:
Cost = 35 cents per ounce * x ounces of Food I + 42 cents per ounce * y ounces of Food II
=> Cost = 35x + 42y (equation 4)
Now, we have the following system of equations:
Equation 1: x + y ≥ 400 / x + 420 / y ≥ 400
Equation 2: x + 3y ≥ 175
Equation 3: 2x + 3y ≥ 280
Objective Function: Cost = 35x + 42y
To solve this system of inequalities and find the minimum cost, we can use graphical or algebraic methods, such as the Simplex method, linear programming, or graphing the inequalities on an xy-coordinate plane.
Note: The units of calcium, iron, vitamin A, and cost are not specified in the question, so we assume they are arbitrary units for the purpose of this problem.
To find the combination of two foods that meets the given nutritional requirements at the minimum cost, we can use a linear programming approach.
Let's assign variables to represent the number of ounces of each food. Let x be the number of ounces of Food I and y be the number of ounces of Food II.
We need to minimize the cost, so our objective function is:
Cost = 35x + 42y
Subject to the following constraints:
Calcium: 1x + 2y ≥ 400 (at least 400 units)
Iron: 2x + 3y ≥ 175 (at least 175 units)
Vitamin A: 3x + 2y ≥ 280 (at least 280 units)
We also have non-negativity constraints:
x ≥ 0 and y ≥ 0
Now, we can solve this linear programming problem to find the optimal values for x and y.
One way to solve this problem is using graphing, which I will explain step by step:
Step 1:
Plot the feasible region by graphing the equations formed by the constraints.
The calcium constraint equation is: x + 2y = 400
The iron constraint equation is: 2x + 3y = 175
The vitamin A constraint equation is: 3x + 2y = 280
Plotting these equations on a graph will give you three lines.
Step 2:
Shade the region that satisfies all the constraints. This region represents all possible combinations of x and y that meet the given requirements.
Step 3:
Calculate the value of the objective function (Cost) at each corner point of the shaded region.
To do this, plug in the values of x and y into the objective function equation (Cost = 35x + 42y), for each corner point of the shaded region.
Step 4:
Compare the values obtained in step 3. The combination of x and y that gives the minimum cost will be the optimal solution to the problem.
That's how you can find the combination of the two foods that gives the diet meeting the requirements of minimum cost.