In Exercises 1–4, use the steps shown in Exercise 1 to find non- negative numbers x and y that s

In Exercises 1–4, use the steps shown in Exercise 1 to find...

Key Concepts

Differentiation

Differentiation is a fundamental tool in calculus used to determine the rate at which a function is changing at any given point. In the context of optimization, differentiation helps identify critical points where the function may achieve a maximum or minimum value by setting the derivative equal to zero.

Critical Point Analysis

Critical point analysis involves finding points where the first derivative of a function equals zero or does not exist. These points are potential candidates for local maxima or minima. Further examination using the second derivative or other tests helps determine the nature of these critical points, which is central to solving optimization problems.

Optimization with Constraints

This concept refers to the process of finding the maximum or minimum of a function subject to certain restrictions. In many practical problems, such as maximizing a product or minimizing cost, the decision variables are limited by constraints that need to be satisfied, making this an essential part of optimization theory in operations research and calculus.

Substitution Method

The substitution method is used to simplify optimization problems by expressing one variable in terms of another using the given constraint. This reduces a multi-variable problem into a single-variable problem, making it more straightforward to find the optimum value using derivative-based methods.

Transcript

00:01 We know that x plus y equals 105 and we are trying to maximize a function and the function we are trying to maximize is x y squared and i'm going to call that function p.

00:22 So p equals x y squared.

00:28 Now the first thing i want to do is solve this first equation for y.

00:35 So i'll have y equals 105 minus x.

00:42 Now we're going to plug that in for y into the p here.

00:49 So we'll have p equals x times 105 minus x.

01:01 Squared.

01:04 Now let's go ahead and determine the domain.

01:07 I'm going to put the domain in red over here.

01:10 So we were told in the problem that x and y were both non -negative numbers.

01:18 So what that tells us is that your x is greater than or equal to zero.

01:24 You also know that your y is greater than or equal to zero.

01:29 Well y is defined in terms of x, so you know 105 minus x is greater than or equal to zero.

01:40 This means that 105 is greater than or equal to x.

01:46 And so if we write that all together, we know that x is greater than or equal to zero, and our x is less than or equal to 105.

01:58 That's our domain.

02:01 Now, we want to take the derivative of p, and i'm going to use the product rule because i have a function times a function, i'll rewrite x, and then i get the derivative of that second function.

02:17 Bring the two in front, and then i have 105 minus x, and the new exponents of 1, so i don't have to write it, and then i'll use the chain rule and get negative 1.

02:30 Plus, now i rewrite the 105 minus x squared and the derivative of the x is 1.

02:43 So what we have is negative 2x times 105 minus x plus, now when i multiply the 105 minus x plus, out? 105.

03:09 I've got to make sure i get all four terms when i multiply that out.

03:19 So when we multiply a binomial times a binomial, make sure that you do 105 times 105, 105 times negative x times 105, and negative x times negative x...

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