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

In Exercises 1–4, use the steps shown in Exercise 1 to find non- negative numbers x and y that satisfy the given requirements. Give the optimum value of the indicated expression.
The sum of $x$ and $y$ is 140 and the sum of the squares of $x$ and $y$ is minimized.

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Key Concepts

Differentiation in Optimization

Differentiation is used to identify critical points of a function by calculating its derivative and setting it equal to zero. This method helps determine where a function attains its minimum or maximum values, which is essential in solving optimization problems.

Constraint Substitution

Constraint substitution is a method used to simplify optimization problems by substituting one variable in terms of another using a given relationship. This reduces a multivariable function to a single-variable function, making it easier to analyze and optimize.

Optimization

Optimization involves finding the best solution (such as a minimum or maximum value) for a problem, often subject to specific constraints. It is widely used to determine the conditions that yield the most favorable outcome for a given function or situation.

Quadratic Function Minimization

This concept focuses on finding the minimum value of a quadratic function, which typically appears as a parabola. When the quadratic coefficient is positive, the function has a single minimum point, which can often be determined through techniques such as completing the square or using calculus.

Transcript

00:01 Okay, the first thing we're told is that x plus y equals 140.

00:11 And we're also told that we want to minimize the sum of the squares.

00:16 So i'm going to write an expression, i'm going to let this p be the equation i'm trying to memorize.

00:23 I want to minimize, minimize the sum of the squares.

00:29 So there is the sum of the squares of x and y.

00:35 Now i'm going to begin by solving x plus y equals 140 for y.

00:41 So i'll subtract x from both sides, and i have y equals 140 minus x.

00:55 Now let's write p in terms of just x.

01:00 So p of x equals x squared.

01:05 Plus 140 minus x squared.

01:18 So in place of the y, i put 140 minus x.

01:21 Now when we simplify this, we get x squared plus 19 ,600 minus minus 280x plus x squared.

01:47 Notice this part right here the 140 minus x squared.

01:55 When you multiply that out you get these three terms.

02:00 Now when we simplify the p of x equals 2x squared minus 280x plus 19 ,000 now we're trying to minimize this, so i'm going to take the derivative and i'm using a different notation than the notation i used in problem one.

02:36 I'm writing p prime of x.

02:38 I could have also written dp over d x.

02:45 So we have two different notations...

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