5.A 1D Optimization

Activities

Characterize the Extrema

  1. Each of the functions below is defined on \((-\infty, \infty)\). Find all the local extrema and global extrema.

image: https://openstax.org/books/calculus-volume-1/pages/4-3-maxima-and-minima

  1. Each of the functions below is defined on a finite interval. Find all the local extrema and global extrema.

image: https://openstax.org/books/calculus-volume-1/pages/4-3-maxima-and-minima

First Derivative Test

Find the critical point(s) of the following functions. Then use the first derivative test to determine whether it is a local minimum, a local maximum, or neither

  1. \(f(x) = x - e^x\)
  2. \(g(x) = x + \sin(x)\) on \([0,2\pi]\)

Second Derivative Test

The function \(f(x) = x^4-8 x^3+22 x^2-24 x\) has critical points at \(x=1\) and \(x=2\) and \(x=3\).

  1. Find \(f''(x)\).
  2. Use the second derivative test to determine whether each of the critical points is a minimum or a maximum.

Cupcake Store

A cupcake store finds that at a price of $4.00, the demand is 400 cupcakes. For every $.25 decrease in price, the demand increases by 20 cupcakes.

  1. Define a function \(q(x)\) that gives the demand for cupcakes at a price of \(x\) dollars.
  2. The revenue for a price of \(x\) dollars is \(r(x) = x \cdot q(x)\). Use your function from part (a) to find the price that maximizes revenue.

Find the price and quantity that maximize the revenue.

Solutions

Characterize the Extrema

  1. We have three different functions
  • \(f(x) = x^3\) has no extrema
  • \(f(x) = \frac{1}{x^2+1}\) has a global maximum at \(x=1\)
  • \(f(x) = \cos(x)\) has global maxima at \(x=2k \pi\) for every integer \(k\). It has global minima at \(x=(2k+1) \pi\) for every integer \(k\).
  1. Once again we have three functions
  • LEFT: global maximum at \(x=0\); local maximum at \(x=4\). Note that \(x=2\) is not a local minimum.
  • MIDDLE: local maximum at \(x=1\); global minimum at \(x=2\); global maximum at \(x=4\).
  • RIGHT: global minimum at \(x=0\). Note that \(x=2\) is not a global maximum (because the function is not defined at \(x=2\)).

First Derivative Test

Find the critical point(s) of the following functions. Then use the first derivative test to determine whether it is a local minimum, a local maximum, or neither

  1. We have \(f(x) = x - e^x\) and so \(f'(x) = 1 - e^x\). There is one critical point: \[ \begin{array}{rcl} 1-e^x &=& 0 \\ e^x &=& 1 \\ x &=& 0. \end{array} \] Using the first derivative test, we have \(f'(-1) = 1 - e^{-1} > 0\) and \(f'(1) = 1 - e < 0\). Therefore \(x=0\) is a local maximum.

  2. We have \(g(x) = x + \sin(x)\) and therefore \(g'(x) = 1 + \sin(x)\). There is one critical point: \[ \begin{array}{rcl} 1 + \cos(x) &=& 0 \\ \cos(x) &=& -1 \\ x &=& \pi \end{array} \]

We have \(g'(\pi/2) = 1 > 0\) and \(g'(3 \pi/2) = 1 > 0\). (In fact, \(1+\cos(x)\) is never negative) Therefore this point is not an extremum.

To find all the extrema of \(g(x)\), we need to consider the endpoints of \([0,2\pi]\). The function never decreases as we move from \(0\) to \(2\pi\).So \(x=0\) is the global minimum, and \(x=2\pi\) is the global maximum.

Second Derivative Test

The function \(f(x) = x^4-8 x^3+22 x^2-24 x\) has critical points at \(x=1\) and \(x=2\) and \(x=3\).

  1. We have \[f'(x) = 4x^3 - 24 x^2 + 44 x - 24\] and \[ f''(x) = 12x^2 - 48x + 44.\]
  2. We check the critical points \(x=1\), \(x=2\) and \(x=3\). \[ \begin{array}{rcl} f''(1) &=& 12 - 48 + 44 = 8 > 0 \\ f''(2) &=& 48 - 96 + 44 = -2 < 0 \\ f''(3) &=& 108 - 144 + 44 > 0 \\ \end{array} \]

Therefore \(x=1\) and \(x=3\) are local minima and \(x=2\) is a local maximum.

Cupcake Store

A cupcake store finds that at a price of $4.00, the demand is 400 cupcakes. For every $.25 decrease in price, the demand increases by 20 cupcakes.

  1. The description of the demand function tells us that \(q(x)\) is a linear function with a negative slope. The slope is \(20/(-.25)=-80\). So \(q(x) = -80x + c\) and we can determine \(c\) by using the fact that \(q(4)=400\). We have \[ 400 = -80 \cdot 4 + c = -320 + c \] so \(c=720\). In conclusion, our demand function is \[q(x) = -80x + 720.\]

  2. The revenue at price \(x\) is \[r(x) = x q(x) = x \left(-80x + 720 \right) = -80x^2 + 720x.\] Taking the derivative and solving for critical points gives \[ -160 x + 720 = 0 \] and therefore \(x=720/160= 4.5\). By the second derivative test, this is a local maximum.

We sell \(q(4.5) = 400 - 40 = 360\) cupcakes and obtain revenue \(r(4.5) = 4.5 \cdot 360 = 1620\).