Maximizing profit and utility functions

[mrjericho1991]

Dear friends, I am new to this forum and I need urgent help in solving these two questions as I am due to submit them tomorrow early morning. Please help me in solving these two questions. Waiting for urgent response.

1: A farm’s profit is given by π = 100x + 80y + 2xy− x(square) − 2y(square) − 5000, where x is the number of turkeys produced and ( is the number of beef cattle produced.
How many of each should be produced to maximise the profit?
Prove that profit is indeed maximised at this level of production?
What is the maximum profit?

2: A consumer’s utility function is given by U(x,y) = xy ( with the budgetary constraint 5x + 10y = 100.

Find the values of x and y ( that maximise the utility function, subject to the budgetary constraint.
What is the value of the Lagrange multiplier λ?

 

[MarkFL]

Hello, and welcome to MHB! (Wave)

I changed the thread title to reflect the nature of the questions being asked. A thread title simply stating help is urgently requested is not useful to the community, nor is it good for SEO. :)

We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?

 

[mrjericho1991]

I have just entered into Economics course and I am going nowhere, I am unable to understand it. I need your complete help step by step in solving these questions.

 

[MarkFL]

I have just entered into Economics course and I am going nowhere, I am unable to understand it. I need your complete help step by step in solving these questions.

Okay, let's look at the first problem. Let's write the profit function as:

\(\displaystyle P(x,y)=100x+80y+2xy-x^2-2y^2-5000\)

We need to set the partial derivatives $P_x(x,y),\,P_y(x,y)$ equal to zero...can you do that?

 

[mrjericho1991]

Not really Sir. I don't want to lie. I want you to solve the question and mention your steps side by side in English. Please assume, I know nothing. Its my first class.

 

[MarkFL]

Not really Sir. I don't want to lie. I want you to solve the question and mention your steps side by side in English. Please assume, I know nothing. Its my first class.

What methods are you given in your lecture notes and/or textbook for maximizing multi-variable functions?

 

[mrjericho1991]

What methods are you given in your lecture notes and/or textbook for maximizing multi-variable functions?

Maxima and Minima of Functions of Two Variables

- - - Updated - - -

Something like critical point, second order derivatives test is also mentioned in here.

 

[MarkFL]

Maxima and Minima of Functions of Two Variables
- - - Updated - - -
Something like critical point, second order derivatives test is also mentioned in here.

Yes, how have you been instructed to optimize such functions (with and without constraint)?

In order to find the critical point(s) without constraint, you need to equate the first partials to zero...do you know how to find partial derivatives?

 

[mrjericho1991]

without constraint. Yes I know but tell me what values do I need to input for partial derivatives?

 

[HallsofIvy]

You don't 'input' values- you set the formulas equal to 0 and solve the equations for x and y.

 

[MarkFL]

without constraint. Yes I know but tell me what values do I need to input for partial derivatives?

You need to equate the first partials to zero, and then solve the resulting system. We have:

\(\displaystyle P(x,y)=100x+80y+2xy-x^2-2y^2-5000\)

And so we compute the first partials, and equation them to zero to obtain the system:

\(\displaystyle 100+2y-2x=0\)

\(\displaystyle 80+2x-4y=0\)

So, what critical point do you get from solving the above linear system?

 

[MarkFL]

As a follow-up, I would begin by adding the two equations to eliminate $x$ and we obtain:

\(\displaystyle 180-2y=0\implies y=90\)

Now, substituting for $y$ into the first equation, there results:

\(\displaystyle 100+2(90)-2x=0\implies x=140\)

And so our critical point is:

\(\displaystyle (x,y)=(140,90)\)

Now, in order to use the second partials test for relative extrema, we need to compute:

\(\displaystyle P_{xx}(x,y)=-2\)

\(\displaystyle P_{yy}(x,y)=-4\)

\(\displaystyle P_{xy}(x,y)=2\)

And we find:

\(\displaystyle D(x,y)=P_{xx}(x,y)P_{yy}(x,y)-\left(P_{xy}(x,y)\right)^2=8-4=4>0\)

And so we conclude that the critical point is at the global maximum, which is:

\(\displaystyle P(140,90)=5600\)

For the second problem, we have the objective function:

\(\displaystyle U(x,y)=xy\)

Subject to the constraint:

\(\displaystyle g(x,y)=5x+10y-100=0\)

Using Lagrange multipliers, we obtain the system:

\(\displaystyle y=\lambda(5)\)

\(\displaystyle x=\lambda(10)\)

We find this implies:

\(\displaystyle x=2y\)

Substituting for $x$ into the constraint, we have:

\(\displaystyle 5(2y)+10y-100=0\implies (x,y)=(10,5)\)

We find:

\(\displaystyle U(10,5)=50\)

Picking another point on the constraint, such as $(12,4)$, we the find:

\(\displaystyle U(12,4)=48<50\)

And so we may conclude that:

\(\displaystyle U_{\max}=50\)

We also find that:

\(\displaystyle \lambda=1\)