There are two factors that limit how much he can bake in a week: He only wants to work for 40 hours a week and he only has one o
ven. Suppose that it takes the baker 1hour to prepare a pair of cakes or a gross of cookies (before they are placed in the oven). Since he only wants to work 40 hours a week, his output of pairs of cakes x and his output of grosses of cookies y are constrained by the equation x+y=40. To maximize the profit of the bakery, the first step is to find where the equations for all of the constraints intersect. For the following part, you will look at x+y=40 and y=0, which is also a constraint (specifically a minimum) since the baker cannot make a negative number of cookies.
Parts A and B might seem easier than most problems with linear systems, but in them you will use the basic techniques needed to solve any linear system: adding equations to cancel variables and substituting the value of one variable to find the value of the other.
Part A
One way of solving systems of linear equations is by adding a multiple of one equation to the other. The multiplier for the first equation is chosen so that one of the two variables will cancel out in the sum. What should you multiply the equation y=0 by so that when added to x+y=40 the variable y will cancel out?
Express your answer numerically.
The coefficient of y in the equation x+y=40 is 1. To make the y-terms cancel when the second equation is added, its coefficient must be -1. It has a coefficient of 1 now, so that equation should be multiplied by -1.
(x +y) +(-1)(y) = (40) +(-1)(0)
x = 40 . . . . . . . simplified; y-terms were cancelled
