Cracking The Ap Calculus ab Exam 2018

This tells us that 11 is a critical point of the equation. Now we need to figure out if this is a maximum or a
minimum using the second derivative.

= 16

Because 16 is always positive, any critical value is going to be a minimum. Therefore, the company
should manufacture 11 units in order to minimize its cost.

Example 3: A rocket is fired into the air, and its height in meters at any given time t can be calculated
using the formula h(t) = 1,600 + 196t − 4.9t^2 . Find the maximum height of the rocket and the time at which
it occurs.

Take the derivative and set it equal to zero.

    =   196 −   9.8t

t   =   20

Now that we know 20 is a critical point of the equation, use the second derivative test.

= −9.8

This is always negative, so any critical value is a maximum. To determine the maximum height of the
rocket, plug t = 20 into the equation.

h(20)   =   1,600   +   196(20) −   4.9(20^2 )  =   3,560   meters

The technique   is  always  the same:   (a) take    the derivative  of  the equation;   (b) set it  equal   to  zero;   and

(c) use the second  derivative  test.

The hardest part of these word problems is when you have to set up the equation yourself. The following
is a classic AP problem:

Example 4: Max wants to make a box with no lid from a rectangular sheet of cardboard that is 18 inches
by 24 inches. The box is to be made by cutting a square of side x from each corner of the sheet and folding
up the sides (see figure below). Find the value of x that maximizes the volume of the box.