Assume that relative maximum and minimum values are absolute maximum and minimum values. Repeat

Assume that relative maximum and minimum values are absolute...

Key Concepts

Multivariable Optimization

This concept involves finding optimal values (maxima or minima) of functions that depend on more than one variable. It typically requires the computation of partial derivatives and solving a system of equations to locate critical points. In economic contexts, this is a common approach for determining equilibrium or optimal pricing and output decisions when multiple interacting factors are present.

Relative and Absolute Extrema

In optimization, relative (or local) extrema are points where a function reaches a maximum or minimum in a small neighborhood, while absolute (or global) extrema are the highest or lowest values over the entire domain. In some problems, assuming that all relative extrema are also absolute extrema simplifies the analysis by allowing one to consider only the critical points found via standard optimization techniques without separately checking endpoint behavior or other regions.

Systems of Linear Equations

Systems of linear equations arise frequently in models where several linear relationships interact. Solving these systems—using techniques such as substitution, elimination, or matrix methods—is crucial for finding the exact values of the variables that satisfy all of the given conditions. This approach is essential in various optimization problems where the equilibrium is determined by the intersection of multiple linear constraints.

Transcript

00:01 This is the same question as question number 20 just for the different numbers.

00:05 So first we'll find the revenue revenue we know that it will be p1 q1 plus p2 q2.

00:12 So that will be 64 we multiply p1 here and we multiply p2 here and then add both.

00:18 So we have 64 p1 minus 4 p1 square minus 2 p1 p2 plus 56 p2 minus 2 p2 minus 4 p2 square.

00:31 This can be rewritten as 64 p1 plus 56 p2 minus 4 p1 square minus 4 p2 square minus 4 p2 so this is the revenue in terms of p1 and p2 rb talks about the values of p1 and p2 and the revenue would be maximum so we find r p1 that is a differentiating with respect to p1 we get 64 fleeting p2 as a constant so we have 64 minus 8 p1 minus 4 p2 and we get r p2 as 56 minus 8 p2 minus 4 p1.

01:16 If equate r p to 0 we have 64 minus 8 p1 minus 4 p2 is equated to 0.

01:23 We divide by 4 so we get 16 is equal to 2 p1 plus p2.

01:29 Let's call this without boxing it up.

01:33 Let's call it equation 1.

01:35 And when our p2 is equated to 0, we have 56 minus 8, p2 minus 4, p1 is equated to 0.

01:43 This means that if we divide by 4, we get 14 as equal to 2p2 plus p1.

01:49 Let's call it equation 2.

01:50 So from equation 1, we get the value of p2 as 16 minus 2 p1, which we place over here.

01:58 So this comes out as 16 is equal to 2 times 16 minus 2 p1 plus 4.

02:04 So this comes out as 14 is 32 minus 3 p1.

02:10 It means that the p1 is 32 minus 14 is 18 and 18 over 3 is 6...