Today we talked about how to find the absolute maximum and minimum of a function on an interval. There are two steps:
Find the critical points (where \(f'(x) = 0\) or does not exist).
Plug the critical points and endpoints into the original function to find the highest and lowest y-values.
We did the following examples in class.
Find the absolute max and min of \(y = 3x^4 - 4x^3\) on \([-1, 2]\).
Find the absolute max and min of \(g(x) = \dfrac{2x}{x^2+1}\) on \([-2,2]\).
Find the absolute max and min of \(y = (x-1)(x-5)\) on \([0,4]\).
Find the critical points of \(y = \sqrt[3]{1+x^3}\).
Find the absolute max and min of \(f(x) = 2x - 3x^{2/3}\) on \([-1,1]\). This one is tricky, because it is easy to miss that 0 is also a critical point! Another good example of the same issue would be to find the absolute max & min of \(y = x^{2/3}\) on the interval \([-1,\tfrac{1}{2}]\).
Some other good examples that we didn’t have time for were:
\(y = x^3 - \frac{3}{2}x^2 - 6x\) on \([-2,3]\)
\(f(x) = x^3(x-4)\) on \([-1,4]\).
Today we talked about optimization word problems. We discussed these additional steps to get started:
Draw and label a picture
Find a formula for the thing you want to optimize.
Find a formula for any constraints, and use it to get rid of extra variables.
We also talked about using the second derivative to test whether a critical point is a max or a min. This is called the Second Derivative Test.
We did the following examples:
An art gallery has 50 prints of works by famous artists to sell for a fundraiser. They estimate that the quantity of prints that people will want to buy will be \(q=500-2p\) when the price of a print it \(p\).
What price should they pick if they want to make sure the quantity demanded is not greater than the 50 prints they have?
What price would be so high that demand would drop to zero?
What price should they sell to raise the most money? This one is tricky because the critical point is outside the domain of the revenue function. Therefore the maximum occurs at an endpoint, not the critical point.
Suppose that cost of running a factory is \(C(x) = 160 + 2x + \frac{1}{10} x^2\), where \(x\) is the level of production. The average cost is \(\dfrac{C(x)}{x}\). What level of production minimizes the average cost? We did this problem two ways, first we took the derivative using the quotient rule which worked but was messy. Then we did the problem again, but using algebra we simplified the formula for average cost to \(160 x^{-1} + 2 + \frac{1}{10} x\), which made it much easier!
Today we looked at other applications of the derivative.
The tangent line of a function is a good approximation for the function close to the point of tangency. We call this the linear approximation function and a formula for it is: \[L(x) = f(x_0) + f'(x_0)(x-x_0).\] We solved the following problems.
Find the linear approximation for \(f(x) = \sqrt{x}\) at \(x_0 = 4\), and use it to estimate \(\sqrt{5}\) and \(\sqrt{3}\).
Find the linear approximation for \(y = \dfrac{360}{x^2}\) at \(x_0 = 10\), and use it to estimate \(\dfrac{360}{9.5}\).
In economics, a demand curve is usually a function \(Q = Q(p)\) that outputs the quantity of a good that will be demanded \(Q\) at a given price \(p\). The price elasticity of demand is \[E = \left| \frac{p Q'}{ Q} \right|\] We say that
The value of \(E\) is the amount that raising the price by 1% will cause demand to fall (as a percent).
We also used calculus to explain the following two facts from microeconomics.
Average cost (\(AC\)) is minimized when average cost equals marginal cost (\(AC = MC\)).
Profit is maximized when marginal revenue equals marginal cost.