SOLUTION: A florist is filling a large order for a client. The client wants no more than 300 roses in vases. The smaller vase will contain 8 roses and the larger vase will contain 12 roses.

Question 1104987: A florist is filling a large order for a client. The client wants no more than 300 roses in vases. The smaller vase will contain 8 roses and the larger vase will contain 12 roses. The client requires that there are at least twice as many small vases as large vases. The client requires that there are at least 6 small vases and no more than 12 large vases.
Let x represent the number of small vases and y represent the number of large vases.

What constraints are placed on the variables in this situation?
Found 2 solutions by josgarithmetic, Theo:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
x small vases
y large vases
x and y must be whole numbers.
Neither x or y is allowed to be more than 300.
300 roses
8 roses per vase, small
12 roses per vase, large

Only analyzed and wrote algebraic statements, but did not solve.

Simple way is start with x at 6, solve for y, and increment x upward by 1 and repeat solve for y. Take the solutions that work.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
let x = number of small vases.
let y = number of large vases.

there are 8 roses in a small vase.
there are 12 roses in a large vase.

your constraints and their associated equations are shown below:

there should be no more than 300 roses total.

8x + 12y <= 300

there need to be at least twice as many small vases as large vases.

x >= 2y

there are at least 6 small vases and no more than 12 large vases.

x >= 6
y <= 12

unspoken, but there:

the number of small vases and large vases must be greater than or equal to 0.

y >= 0

x >= 0 not required since x >= 6 takes care of that.

summary of constraint equations:

8x + 12y <= 300
x >= 2y
x >= 6
y <= 12
y >= 0

if you graph the opposite of these inequalities using the desmos.com calculator, the unshaded area of the graph will be your feasibility region.

that graph is shown below.

the maximum or minimum value of the objective function will be at the corner points of the feasible region.

since this problem doesn't have an objective function, the value at those corner points is not used.

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