Question

Each week, a manufacturer of chocolate cakes is able to produce f(H, M) cakes if their employees work a total of H hours and they use their cake production machines for a total of M hours. The employees’ wage is $19 per hour and due to electricity and maintenance, the machines cost about $30 per hour to run. Their operating budget is $10,000 per week. In order to determine how to maximize their production, they use the method of Lagrange multipliers. The solutions to the Lagrange multiplier equations are H = 400, M = 80, and λ = 5. (a) (3 points) If their weekly budget decreases by about $100, how will their weekly production be affected?

(b) (3 points) Find the value of fH(400, 80).

(c) (5 points) In order verify that their solution does indeed lead to maximum production, they use the second derivative test. At the point (H, M) = (400, 80), they find that D = fHHfMM −(fHM) 2 > 0 and fHH > 0. Should they be concerned? Please explain.

Answer 1

Answer:

Given that,

Each week, a manufacturer of chocolate cakes is able to produce f(H, M) cakes if their employees work a total of H hours and they use their cake production machines for a total of M hours.

The employees’ wage is $19 per hour and due to electricity and maintenance, the machines cost about $30 per hour to run.

Their operating budget is $10,000 per week. In order to determine how to maximize their production, they use the method of Lagrange multipliers.

The solutions to the Lagrange multiplier equations are H = 400, M = 80, and λ = 5.

Kindly go through the solution provided below.

One thing to notice here is that 1 week = 7 days = 7*24 hours = 168 hours. Therefore it is not possible for the workers to work for 400 hours! in a week.

Nevertheless,

Production f(H,M)

Total Cost=19H+30M

Operating budget=10,000

Constraint Total cost Operating budget

19H+30M 10,000

Let g(x)=19H+30M-10,000

Then, for g(x)=0, we can model the given problem using the method of Lagrange multiplies.

(a).

If the weekly budget decreases by $100, then

g(x)=19H+30M-9,900

i.e, H=400, M=80 will no longer be the solution to the problem.

The new solution will be H=390, M=83

Now, we know from the Taylor series expansion for a two-variable function that,

for h=-10; k=+3 we can write,

=f(400,80)-10(95)+3(150)

=f(400,80)-500

We can see that f(390,83) < f(400,80)

Therefore, we conclude that the production will decrease.

(b).

, as calculated in (a).

(c).

From the second derivative test if D>0 and , then f(H,M) has a local minimum at (H,M).

In other words for the given budget constraint, the production is very low. And the production can be increased by operating at a lower cost.

That is they are not producing at an optimal level. It is possible that they have redundant labor.