Have no explicit solution in terms of elementary functions. Use a CAS to explore graphically eac

Have no explicit solution in terms of elementary functions....

Key Concepts

Differential Equations

Differential equations describe relationships between a function and its derivatives, capturing how a quantity changes over time or space. They are foundational in mathematics and many applied fields, enabling the modeling of dynamic systems in physics, biology, economics, and engineering.

Non-Elementary Solutions

Not all differential equations can be solved explicitly using elementary functions. Sometimes, the solution exists in a form that is not expressible in closed-form expressions composed of basic operations and well-known functions, necessitating numerical or graphical methods for analysis.

Gompertz Equation

The Gompertz equation is a specific type of differential equation used to model growth processes, particularly in biological contexts such as tumor growth or population dynamics. It features a logarithmic term that regulates growth, representing scenarios where growth slows as the system approaches a limiting value.

Computer Algebra Systems (CAS)

Computer Algebra Systems are software tools designed to perform symbolic mathematics, including the manipulation and solution of differential equations. They are particularly useful for exploring equations that do not have explicit solutions in terms of elementary functions, as they provide graphical and numerical methods to analyze the behavior of such systems.

Transcript

00:01 Function and we are asked to find the absolute max and minima of this function with the given domain.

00:06 So here we are told that x is between the values of zero and one and that y is also between the values of zero one.

00:13 So here we're going to have this box that we are looking at.

00:20 And i'm going to go ahead and label my vertices.

00:25 So to go about this problem, there are several places where these absolutely extreme may can happen.

00:32 So first, i'm going to go about this problem.

00:32 Look at my boundary points, and then i'm going to look at my interior points and see where i get my biggest and smallest number.

00:40 And we're going to do the same process that we've done in previous problems.

00:45 So i'm going to first look at my line segment, oa.

00:53 So looking at that, my x value is fixed at zero, so it's going to be the same thing as f of 0y.

01:00 And so when i plug in 0 for x into this equation, i get negative 24 y squared.

01:07 And now just like before, to get our extreme value i'm going to take the derivative and set it equal to zero.

01:15 So when i take the derivative of this i get negative 48 y and so therefore y is equal to zero and x is equal to zero.

01:27 However, that was already one of my endpoints.

01:30 So now i'm just going to find the values of my endpoints so that's also where extreme values can happen.

01:39 So when i plug in zero into my function i just get zero and then when i plug in zero for x and one in for y i get negative 24 okay then i'm just repeat this process with all of the lines like all of the side lengths so for ab now y is fixed at 1 x is with the one that's changing so my f of x y is going to equal to f of x1 so plugging in one for y into this equation i end up with 40 x minus 32 x cubes minus 24 and then taking the derivative of this i get 48 minus 96 x squared is equal to zero and so if you solve that for x you should get the x is equal to plus or minus one over the square root of two or y is equal to one so these are our points or interior points of that line however, the negative is not an interior point.

02:56 It's not in our domain, so we're going to forget the negative and just finds the value of the positive, and then also my endpoints.

03:07 I already found this end point, so i just need to find it.

03:11 F of 1 .1.

03:13 So the value of this function at this point is going to be 16 times the square root of 2 minus 24, and at 1 -1, it's going to be negative 8.

03:27 Okay, moving on, i'm going to go to bc.

03:32 So here, my x is fixed at 1.

03:37 Y is the one that's changing.

03:38 So it's going to be the same thing as f of 1.

03:42 Or 1y, sorry.

03:47 And when i plug in 1 for x, i get 48.

03:52 Y minus 32, minus 24 y squared...