Among all pairs of numbers whose difference is 14, fi nd a...

Key Concepts

Completing the Square

This algebraic method transforms a quadratic expression into a perfect square form plus a constant, thereby making it straightforward to identify the vertex of the parabola. This technique is particularly useful in optimization problems as it directly reveals the minimum or maximum value of the function in question.

Minimization of Quadratic Functions

The process of finding the minimum value of a quadratic function involves identifying its vertex, where the function reaches its lowest value for an upward-opening parabola. Techniques such as completing the square or applying the vertex formula (derived from the coefficients of the quadratic) are commonly used to determine this critical point.

Quadratic Function Formation

When a product or other function is expressed in terms of a single variable through substitution, it often leads to a quadratic function. Quadratic functions are fundamental in algebra and optimization because their parabolic shape guarantees a single minimum or maximum point under standard conditions.

Constraint-based Substitution

This concept involves using given conditions to express variables relative to each other. In optimization problems, a constraint (like a fixed difference between two quantities) can reduce the number of independent variables, allowing the expression of the function to be optimized in a single variable.

Transcript

00:01 Okay, among all pairs of numbers whose difference is 14, find a pair whose products is as small as possible.

00:07 What is the minimum product? okay, so a big or word of question here.

00:12 Let's get some equations down, shall we? so the difference between our two pairs of numbers is 14.

00:19 So one minus the other should equal 14.

00:23 I hope you can see that.

00:24 I hope you see where i've got that from.

00:26 And we want to minimize x, 1.

00:30 The product of these two we want to find the minimum the minimum value here we'll call it z for now all right we'll call it z okay now how let's let's get this first equation so that we can use yeah so totally y across minus the 14 y is equal to x minus 14 okay we want to minimize this equation however we've got three unknowns right so let's substitute our y value into here okay so you have x lots of x and x lots of minus 14 so we have x square minus 14 x is equal to our unknown we want to get a minimum value for this right and how do we find minimums well we all we need to do is differentiate and evaluate at zero because our minimum point will be when the gradient is zero okay i hope you know that from background stuff.

01:36 I haven't gone too into depth for why that works or anything.

01:39 But i hope you can see that this will be our minimum point for this equation here.

01:43 Yeah.

01:44 So let's differentiate.

01:45 Times by the power and take one away.

01:47 So we've got two lots of x.

01:51 And we've got one lot of minus 14.

01:55 Okay...