Extrema of functions of two variables

In summary, the conversation discusses the process of maximizing profit for a corporation that manufactures candles at two locations. The cost of producing units at each location is given by separate equations, and the selling price is fixed at $15 per unit. The conversation mentions using "x" and "y" instead of "x1" and "x2" for convenience, and taking partial derivatives to find the maximum profit. However, the correct method is to find the intersection of the two cost equations to determine the ideal production quantities at each location.
  • #1
Math87
1
0

Homework Statement



A corporation manufactures candles at two locations. The cost of producing x1 units at location 1 is C1= 0.02X12+4X1+500
and the cost of producing x2 units at location 2 is C2=0.05X22+4X2+275.

The candles sell for $15 dollars per unit. Find the quantity that should be produced at each location to maximize the profit.

p=15(X1+X2)-C1-C2


The Attempt at a Solution


First off, my professor said "This is just for convenience, especially when plotting it in wolfram alpha, you can't enter x1 and x2 but you can enter x and y. So in the give equations, change the letters, making x1 into x and make x2 into y. Then you will have an equation with numbers and x and y ."
i changed everything to x and y

C1= 0.02X2+4X+500
C2= 0.05Y2+4Y+275
Then i decided to combine them together to get:


0.02X2+4X+500-0.05y2-4Y-275=0
After that i found my 6 partial derivatives, but i have a feeling I am doing this all wrong..
Can you give me tips on how to do this question please.
 
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  • #2
At the end of your post you wrote down the equation [tex] C_1 - C_2=0[/tex]
and then started taking partial derivatives of the left hand side.

This has basically nothing to do with the actual objective: maximizing [tex]p(x,y)=15(x+y)-C_1-C_2 = 15(x+y)-.02x^2-4x-500-.05y^2-4y-275[/tex]

Given a function, how do you find its maximum?
 
  • #3
I hope I'm not doing this wrong :D
I got few things to say :)
First,you actually can enter values with the name 'x1' and 'x2'. Give it a try :)
Second,in your equation ,where you substituted 'x1' and 'x2' with X and Y.In Mathematica it's better to use lowercase letters for your functions,variables,constants,etc.Some of the uppercase letters are actually preset by the programmers.
Third,the real problem :)
So to find where you will maximize the profit,you should be looking for the lowest budget for producing the candles - the intersection of the two graphics.Finding it by hand it's relatively easy,but you can also use Mathematica.
3.png

So the quatity C1=992 and C2=1015,and the profit p(x)=573
Hope I solved that right :biggrin:
 

What are extrema of functions of two variables?

Extrema of functions of two variables are the highest and lowest points on the surface of a three-dimensional graph. They represent the maximum and minimum values of the function in the given domain.

How do you find extrema of functions of two variables?

To find the extrema of a function of two variables, you must first take the partial derivatives of the function with respect to each variable. Then, set both partial derivatives equal to zero and solve for the variables. The resulting points will be the extrema.

What is the difference between a relative and absolute extremum?

A relative extremum is a point on the graph where the function changes from increasing to decreasing or vice versa. An absolute extremum is the highest or lowest point on the graph, regardless of the direction of change.

Can a function have multiple extrema?

Yes, a function of two variables can have multiple extrema. It can have both a maximum and minimum value, or multiple points of either type.

How are extrema of functions of two variables used in real life?

Extrema of functions of two variables are used in many real-life applications, such as optimization problems in economics, engineering, and physics. They can also be used to analyze the behavior of natural phenomena, such as weather patterns or population growth.

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