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AP Calculus Question of the Month: July
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by Ben Klein
Davidson College
Davidson, North Carolina
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| | The Area Problem in 2004's Form B AB6, BC6
Problem 6 on the 2004 AB and BC Form B examinations was a common problem with three parts, the first two of which were essentially hints to help students solve the third part of the problem. Even though you can download the problem for yourself, we will reproduce it here along with an unofficial solution on the Answers Page. We will then introduce a similar problem and invite you to solve it.
AB/BC6
Let 
be the line tangent to the graph of 
at the point (1,1), where
, as shown above.
(A) Find 
.
(B) Let T be the triangular region bounded by
, the 
-axis, and the line
= 1. Show that the area of T is 
.
(C) Let S be the region bounded by the graph of
, the line
, and the
-axis. Express the area of S in terms of 
and determine the value of
that maximizes the area of S.
The original AB/BC6 ended here. You may wish to work these original parts first and then check your answers before going on to the following new problems. The first of the new parts is directly related to part (C), but the remaining new parts will ask you to consider a situation that is similar to the original one but is somewhat more complicated. This time the region in question will be defined by two parameters, not just one.
(D) Verify the formula given in the answer to (C) for the area of S, thought of as the region between two curves, by direct integration. (Note that the "bottom curve" for S is the
-axis for small
but is the line
for large
.)
Now we let
be the line tangent to the graph of
at the point 
where 
and
> 1 . Define a triangular region R, which depends on both 
and
, as follows: the vertices of R are the
-intercept of
, the point (1,0), and the point at which
meets the line
= 1. You should draw a sketch of the region R before going on. Note that if
= 1, the region R is the region T in the original problem.
Now work the following problems. You may want to do the first problem and check your answer before going on to the rest of them.
(E) Let
= 3, and show that the area of R is 
. Next find the value of
that maximizes the area of R, and then find the corresponding maximum area.
(F) Show that for an arbitrary value of
> 1, the area of R is given by 
. Use this to find the value of
that maximizes the area of R and the corresponding maximum area. Your answer should depend on the parameter
.
Since the region R is contained in the region bounded above by the graph of
, below by the
-axis, and on the right by the line
= 1, the area of R is less than 
, the answer to part (A). Thus, the maximum area of R, as a function of
, should go to zero as
goes to infinity.
(G) Show by direct calculation that the maximum area of R that you found in part (F) does go to zero as
goes to infinity.
Click here to view the answers and commentary!
Ben Klein is currently the Beverly F. Dolan Professor of Mathematics at Davidson College in Davidson, North Carolina, where he has taught since 1971. Ben's relationship with AP Calculus began in 1990 when he served as a Reader at Clemson University. He has attended every Reading since then and has served as a Table Leader in recent years. He just completed a four-year term on the AP Calculus Development Committee.
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