Alocal bagel shop produces two products: bagels (b) and croissants (c). each bagel requires 6 ounces of flour, 1 gram of yeast, and 2 tablespoons of sugar. a croissant requires 3 ounces of flour, 1 gram of yeast, and 4 tablespoons of sugar. the company has 6,600 ounces of flour, 1,400 grams of yeast, and 4,800 tablespoons of sugar available for today's production run. bagel profits are 20 cents each, and croissant profits are 30 cents each. a) max profit = x +y (use cents) b) subject to: + < = , + < = + < = 4800 sugar x, y > = 0

  see below

Step-by-step explanation:

If we let X represent the number of bagels produced, and Y the number of croissants, then we want ...

(a)   Max Profit = 20X +30Y

(b)   Subject to ...

     6X +3Y ≤ 6600 . . . . . . available flour

     X + Y ≤ 1400 . . . . . . . . available yeast

     2X +4Y ≤ 4800 . . . . . . available sugar

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Production of 400 bagels and 1000 croissants will produce a maximum profit of $380.

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In the attached graph, we have shaded the areas that are NOT part of the solution set. (X and Y less than 0 are also not part of the solution set, but are left unshaded.) This approach can sometimes make the solution space easier to understand, since it is white.

The vertex of the solution space that moves the profit function farthest from the origin is the one we are seeking. The point that does that is (X, Y) = (400, 1000).

a negative exponent is translated as follows:

so, we have