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Here's a simple optimization problem to get us started:

4.7 #2 From Calculus Early Transcendentals 5th Edition

Find 2 numbers whose difference is 100 and whose product is minimum. So we know a very basic information:

a) let the 2 numbers be x and y.

b) y - x = 100 so y = 100 + x

c) we want to minimize xy or since we have y in terms of x, x(100 + x) = x^2 + 100x

Now let's minimize the product, or xy.

f(x) = x^2 + 100x

find the derivative

f ' (x) = 2x + 100 {power rule 2 times}

Find the critical points by setting f ' (x) = 0

if 2x = + 100 = 0, 2x = -100 and x = -50 (critical value).

Now that we have a critical point we need to assess whether or not the function is increasing or decreasing around that point. Let's find values of f prime = f ' (x) for x = -49, -50, -51

• f ' (-51) = -2 -- if the first derivative is less than zero the function is decreasing on its interval.
• f ' (-50) = 0 -- recall that if x = -50 the derivative is zero, which implies that the curve is experiencing a horizontal tangent.
• f ' (-49) = 2 -- so the curve is going up as the first derivative is +.

If the curve is going down at -2, flat at -50 and increasing at 2 onward, I'd say that our critical point is a minimum. \ _-50_ /.

Well - there isn't much left to do but solve for y.

y = 100 + x
y = 100 -50
y = 50

That's that! Next post we'll try a more complex optimization problem.