SOLUTION: The manufacturer of dolls want to maximize their revenue for the sale of the dolls. Each girl doll sold yields $9 in revenue and each boy doll yields $9.50. The dolls are made of p

Question 1123606: The manufacturer of dolls want to maximize their revenue for the sale of the dolls. Each girl doll sold yields $9 in revenue and each boy doll yields $9.50. The dolls are made of plastic and nylon. To make a girl doll requires 12 oz of plastic and 5 oz of nylon. A boy doll requires 14oz of plastic but no nylon. The manufacturer has no more than 100,000 oz of plastic and 30,000o of nylon available each week. Kids will buy at least twice as many girl dolls as boys. How many of each should they manufacture each week to maximize their revenue?
Found 3 solutions by Theo, ikleyn, greenestamps:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
your objective function is 9x + 9.5y which you want to minimize.

your constraint functions are:

12x + 4y <= 400,00 (plastic constraint)
5x <= 30,000 (silk constraint)
x,y >= 0 (can't be negative)

using the desmos.com calculator, you would graph the opposite of these constraints.
the area of the graph that is not shaded is your region of feasibility.

the corner points of this region are where the maximum revenue will be.
you will evaluts the objective function at these corner points to determine thye corner point that provides the most revenue.

you also need to confirm that all constraints are satisfied as the corner point with the maximum revenue.

the graph looks like this:

the corner point with the maximum revenue is at (6000,2000)

the revenue at this corner point was 73,000, which was higher than at any of the other corner points.

the plastic consumed was 100,000 and the nylon consumed was 30,000, all under the constraint limits of <= 100,000 and <= 30,000.

the cost function at the corner point was 9 * 6,000 + 9.5 * 2000 = 73,000

the plastic constraint was 12 * 6000 + 14 * 2000 = 100,000

the nylon constraint was 5 * 6000 + 0 * 2000 = 30,000

the desmos.com calculator can be found at https://www.desmos.com/calculator

Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
Unfortunately, the solution by @Theo contains mistakes.
Some mistakes are technical (like typos).
But there are conceptual mistakes, too.
So, I came to put here the correct solution.

Let x = # of girl dolls; y = # of boy dolls. Your objective function is R(x,y) = 9x + 9.5y which you want to minimize. your constraint functions are: 12x + 14y <= 100000 (plastic constraint) <<<---=== I edited this line 5x <= 30000 (silk constraint) x >= 2y ("Kids will buy at least twice as many girl dolls as boys.") <<<---=== missed constrain x,y >= 0 (can't be negative) Region of feasibility is shown in the plot below. It is the quadrilateral in Quadrant I below the blue line, below the red line and to the left from the vertical green line. // To make the numbers on the coordinate axes better visible, I showed the values along coordinate axes in thousands dolls. Plots y = (red), x = 6 (green) and (blue) The corner points of this region are where the maximum revenue will be. You will evaluate the objective function at these corner points to determine the corner point that provides the most revenue. You also need to confirm that all constraints are satisfied as the corner point with the maximum revenue. The corner point with the maximum revenue is at (6000,2000) The revenue at this corner point was 73,000, which was higher than at any of the other corner points. The plastic consumed was 100,000 and the nylon consumed was 30,000, all under the constraint limits of <= 100,000 and <= 30,000. The cost function at the corner point was 9 * 6,000 + 9.5 * 2000 = 73,000 The plastic constraint was 12 * 6000 + 14 * 2000 = 100,000 The nylon constraint was 5 * 6000 + 0 * 2000 = 30,000

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Further comments on this problem and the previous "combined" solution by @theo and @ikleyn....

@ikleyn points out that @theo overlooked one of the constraints -- although, as it turns out, that constraint doesn't alter the answer @theo got.

My comments are about three statements in their solution:

(1) The corner points of this region are where the maximum revenue will be.

(2) You will evaluate the objective function at these corner points to determine the corner point that provides the most revenue.

(3) You also need to confirm that all constraints are satisfied as the corner point with the maximum revenue.

(1) Poorly stated. The maximum revenue (maximum value of the objective function) will USUALLY be at ONE OF the corner points of the feasibility region. However, in some problems the maximum value of the objective function can be obtained at any point along one edge of the feasibility region.

Note that is equivalent to saying the maximum value will NEVER be obtained at any point IN THE INTERIOR of the feasibility region.

(2) Not true. You don't have to evaluate the objective function at every corner of the feasibility region.

Where the maximum value of the objective function is obtained is exactly determined by the slope of the objective function and the slopes of the constraint boundary lines.

In this problem, the slopes of the constraint boundary lines are 1/2 and -6/7; for this problem, we can think of the slope of the vertical constraint boundary line as "negative infinity".

The slope of the objective function is -18/19, which is more negative than -6/7, and less negative than "negative infinity". That means the maximum value of the objective function will be obtained at the intersection of the boundary constraint lines with slopes of -6/7 and "negative infinity" -- at (6000,2000).

If you have trouble seeing this, imagine all the lines with the slope of the objective function, -18/19. You want the line with that slope that just touches the feasibility region.

That is usually going to be at one of the corners; however, note that if the slope of the objective function is the same as the slope of one of the constraint boundary lines, the maximum value of the objective function can be anywhere along an edge of the feasibility region.

(3) NOT true!! Of course you don't have to check that the constraints are satisfied at the corner point that gives the maximum value of the objective function. All the constraints are, by definition, satisfied at ALL points in, or on the boundary of, the feasibility region.

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