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,...

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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 = *0.6*a + *0.4*b *(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 = *3*a* + 1*b* (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 = *1*a* + 1*b* (1 unit in each can of A and 1 unit in each can of B). The requirement for carbohydrates is

`a+b >=4`

Finally, for fat

F = 2*a* + 4*b *(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 2*a* + 4*b* = 12 and *a* + *b* = 4)

(1, 3) (The intersection of 3*a* + *b* = 6 and a + *b = *4)

(1.2, 2.4) (The intersection of 2*a* + 4*b* = 12 and 3*a* + *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.**