Time The time (in hours) that a branch of Amalgamated Entities needs to spend to meet the quota

Time The time (in hours) that a branch of Amalgamated...

Key Concepts

Second Derivative Test

The second derivative test, often involving the Hessian matrix in multivariable settings, is used to determine the nature of a critical point. By analyzing the eigenvalues or the determinant of the Hessian matrix at a critical point, one can classify the point as a local minimum, local maximum, or saddle point, thereby confirming the type of extremum.

Critical Points

Critical points of a function are points where all the partial derivatives vanish or fail to exist, and they are candidates for local minima, local maxima, or saddle points. Identifying these points is a crucial step in the optimization process as it narrows down the set of potential solutions for where the function may reach its extreme values.

System of Equations

In the context of optimization, setting the partial derivatives equal to zero results in a system of equations that must be solved simultaneously. Solving this system yields the candidate values for the variables that potentially minimize or maximize the function, and often requires algebraic or numerical methods to find the solution.

Multivariable Optimization

Multivariable optimization involves finding the extreme values (minimum or maximum) of functions with several variables. This process requires examining how the function behaves with respect to each variable and identifying points where the function might attain its optimum value, which often involves analyzing both the gradient and the curvature of the function.

Partial Derivatives

Partial derivatives measure the rate of change of the function with respect to one variable while keeping the other variables constant. They are fundamental in multivariable calculus as they help identify how changes in individual variables influence the overall function, making them an essential tool in locating critical points where the function’s slope is zero.

Transcript

00:01 So for this problem, we're asked to minimize this function, t of x of y, where t is total time spent for this company to reach quota.

00:12 X is number of thousands of dollars spent on quality control, and y is spent on consulting.

00:20 So really, this is just a critical points problem.

00:22 So we want to find the critical points of this function t.

00:25 And so we're going to start by doing our partial derivatives, starting with x.

00:32 This is 4x cubed.

00:35 Y is a constant.

00:36 So we're just left with 32y.

00:39 We do the partial derivative with respect to y.

00:43 Now, x is a constant.

00:45 This is a 64y cubed.

00:48 It's 4 times 16.

00:49 40 and 24.

00:50 Yeah, 64 way cubed minus 32x.

00:55 Again, x is just a constant.

00:57 Now we're looking for the points ab that make both of these functions zero.

01:02 So t sub x of ab is equal to zero implies we've got 4a cubed minus 32b equals 0.

01:17 So this implies we're going to use we're going to solve for a in terms of b.

01:22 So this is 4a cubed.

01:24 It's equal to 32b.

01:26 This is going to imply that a cubed is equal to 8b or a is equal to 2 times b to the 1 3rd that's 8 to the 1 3rd that's 2 cubed is 8 so now that we know what a is in terms of b we now look at our other expression t y ab is equal to 0 this implies that c 64 b cubed minus 32 times a and a is 2b to the 1 third is equal to 0 so this is 64 b cubed minus 64 b to the 1 3rd equals 0 we have 64 is a common value between these two terms so we take that out we can also take out a b to the 1 third.

02:34 This leaves us with b to the 8 thirds.

02:37 1 third and 8 thirds makes 9 over 3, which is 3, minus 1 equals 0.

02:47 So now we can see our two solutions.

02:49 B equals 0 is one solution.

02:51 And if we put in b equals 1, that makes this whole term in the brackets 0.

02:57 And so our other solution is a positive.

03:03 We're not going to consider a negative 1.

03:06 Because that would be negative dollars spent, which doesn't really make sense.

03:12 So positive 1.

03:15 All right, we knew from above that a was equal to 2b to the 1 3rd.

03:21 So our values of a when b is equal to 0, a is equal to 0, and when b is equal to positive 1, a is equal to positive 2.

03:31 So here's our two critical points.

03:34 So we've got 0 comma 0, so they don't spend any money on either of those two things.

03:39 And we've also got $2 ,000 spent on quality control and $1 ,000 spent the consulting.

03:50 So now we have to calculate our discriminant, which looks like this, all evaluated at our points ab.

04:02 So we have to calculate these functions, f of x, x, because we want to find out which of the solutions that we found is a minimum, because we're trying to minimize our time.

04:13 We don't want to choose the one that maximizes our time spent, because the company probably won't be very happy with us if we do that.

04:21 So we're going to go back up here, and we're going to just collect these two terms that we already calculated.

04:30 I need to click on here somewhere.

04:36 Paste, there they are.

04:42 I said i wrote f's, but this is really t, x, x, t, sub y, and t sub y, and t sub x, y, and t sub x, y.

04:53 So to calculate this first term here, we're taking the derivative of this term with respect to x.

05:00 So that's going to give us 12x squared...

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