Question

GIVING OUT BRAINLIEST!! All sacks of sugar have the same weight. All sacks of flour also have the same weight, but not necessarily the same as the weight of the sacks of sugar. Suppose that two sacks of sugar together with three sacks of flour weigh no more than $40$ pounds, and that the weight of a sack of flour is no more than $6$ pounds more than the weight of two sacks of sugar. What is the largest possible weight (in pounds) of a sack of flour?

Answer 1

Answer:

flour = 11.5, sugar has to be equal to 2.75 pounds.

Step-by-step explanation:

your 2 inequalities are:

2x + 3y <= 40

y <= 2x + 6

the second equation can be shown as:

-2x + y <= 6

solve the equality portion of these 2 inequalities.

those are the following equations:

2x + 3y = 40

-2x + y = 6

add these equations together and solve for y to get

y = 11.5

that appears to be a break even point.

when y = 11.5, the first inequality is solved as follows:

2x + 3y <= 40

replace y with 11.5 and the inequality becomes:

2x + 34.5 <= 40

solve for x to get

x <= 2.75

when y = 11.5, the second inequality is solved as follows:

-2x + y <= 6

replace y with 11.5 and the inequality becomes:

-2x + 11.5 <= 6

solve for x to get x >= 2.75

this indicates that, when y = 11.5, x has to be equal to 2.75.

both constraints are satisfied.

Answer 2

Answer:

Let each sack of sugar weigh $s$ pounds and each sack of flour weigh $f$ pounds. Then from the given information, $2s + 3f \le 40$ and $f \le 2s + 6.$ Also, $s > 0$ and $f > 0.$ We graph these the intersection of these inequalities below.(document attached

We see that the largest possible value of $f$ occurs at the points where the lines $2s + 3f = 40$ and $f = 2s + 6$ intersect. Solving for $f,$ we find $f = \boxed{\frac{23}{2}}.$

Step-by-step explanation: