This is a kind of optimization problem, and it can be solved by using a system of inequalities. Let's denote the number of cans of brand A dog food by a and the number of cans of brand B dog food by b. The goal is to minimize the cost, which can be expressed, in dollars, as
C = 0.6a + 0.4b (60 cents for each can of A and 40 cents for each can of B)
The constraints on a and b come from the minimum requirements for each nutrient. The amount of protein in a cans of brand A dog food and b cans of brand B dog food is
P = 3a + 1b (3 units of protein in each can of A and 1 unit of protein in each can of B). According to Mr. Smith, the amount of protein has to be at least 6 units a day, so
`3a+ b >=6` .
Similarly, the amount of carbohydrates in the same combination of two brands will be
CH = 1a + 1b (1 unit in each can of A and 1 unit in each can of B). The requirement for carbohydrates is
Finally, for fat
F = 2a + 4b (2 units in each can of A and 4 units in each can of B). The minimum amount of fat is 12 units:
`2a + 4b >=12`
The three constrains obtained above are graphed below. (a is graphed on the horizontal axis, and b is graphed on the vertical axis.) The region formed by the intersecting lines is a triangle with the vertices
(2, 2) (The intersection of 2a + 4b = 12 and a + b = 4)
(1, 3) (The intersection of 3a + b = 6 and a + b = 4)
(1.2, 2.4) (The intersection of 2a + 4b = 12 and 3a + b = 6)
(These results can be verified by solving the corresponding systems of equations.)
The cost (or any function of a and b) will have its maximum and minimum values at the one of the vertices of the triangle. (This is a rule for optimization of a linear function with linear constraints.)
`C(1, 3) = 0.6*1 + 0.4*3 = 1.8`
`C(2,2) = 0.6*2 + 0.4*2 = 2`
`C(1.2, 2.4) = 0.6*1.2 + 0.4*2.4 = 1.68`
The least of these values is 1.68 dollars, which is the cost of 1.2 can of brand A dog food and 2.4 cans of brand B dog food.
So, Mr. Smith should feed his dog 1.2 can of brand A dog food and 2.4 cans of brand B dog food each day. This will cost him $1.68 a day, which will be the least amount of money he can spend and still satisfy the nutritional requirements.