where x1 and x2 represent number of units of production of basic and advanced iPods, respectively. Production time required for the basic iPod is 6 hours per unit, and production time required for the advanced iPod is 8 hours per unit. Currently, 50 hours are available. The cost of hours is already factored into the profit function. Formulate an optimization problem that can be used to find the optimal production quantity of basic and advanced iPods. Solve the optimization model you formulated, what is the optimal value of the objective function? Round your answer to one decimal place.

Optimization function is a real-valued function that is either minimized or maximized relative to the given set of feasible variables. Therefore if our maximum value for the production of both types of iPod is 50 hours and if it takes 6 hours to produce basic iPod and 8 hours to produce advanced iPod than optimization model is as follows 6 *x + 8*y = 50. From there we have a set of possible solutions:

if x=0 than y= 6,2 , which is not possible as it is not the natural number
if y=o than x=3,57, which is also not natural number
the only solution which satisfy the above condition is where x=3 and y=4

sqdancefan sqdancefan · Answer 2 · 2020-04-25T21:04:44+02:00

Answer:

maximize P(x1, x2) = (x1)(38 -5x1) +(x2)(44 -5x2) +520 subject to 6x1 +8x2 ≤ 50
(x1, x2) = (3, 4)
P(3, 4) = 685

Explanation:

Given

x1, x2 are units of production

6x1 +8x2 ≤ 50 . . . . limit on production hours

P = x1(38 -5x1) +x2(44 -5x2) +520 . . . . profit function

Find

a formulation for the optimization problem to maximize profit

the solution to that problem

maximum profit

Solution

There are a number of ways to approach this problem. One is to use the method of Lagrange multipliers. Another is to use the hour constraint to find x2 in terms of x1, then formulate the profit maximization as a problem in one variable. A graphical approach is another possibility.

For simplicity here, we'll try the second listed approach.

x2 = (50 -6x1)/8 = (25 -3x1)/4

Then the optimization problem is ...

maximize: P(x1) = x1(38 -5x1) +(25 -3x1)/4(44 -5/4(25 -3x1)) +520

for integer values of x1 and (25-3x1)/4

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P(x) can be simplified to ...

P(x) = (-5/16)(25x^2 -166x -1919) = (-125/16)(x^2 -6.64x -76.76)

Converting this to vertex form, we have ...

P(x) = (-125/16)((x -3.32)^2 -65.7376)

The value of x1 that will optimize profit is x1 = 3.32. The corresponding value of x2 is (25-3×3.32)/4 = 3.76.

These are not integers, so we need to find integer values of x1 and x2 that satisfy the hour constraint and also maximize profit. We can start with ...

for (x1, x2) = (3, 4) . . . . hours needed are 6(3) +8(4) = 50 (OK)

profit is 3(38 -5×3) +4(44 -5×4) +520 = 685

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We can verify this result using a graphical approach.

We find the profit function describes a circle with center (x1, x2) = (3.8, 4.4) and that maximizing profit is equivalent to finding the circle of smallest radius. Of course, the circle must pass through a grid point that is in the feasible region defined by the hours limit. The closest grid point to (3.8, 4.4) is (3, 4).