Question 83815: An professor finds that the number of students enrolling in his class is approximated by s(t) = -tsquared + 20t + 80 where t is the number of tests given in one semester.
(a) How many tests should the professor give so that the number of students enrolling in the class is a maximum?
(b) If the professor does not want students to enroll, how many tests should he give?
Found 2 solutions by stanbon, Edwin McCravy:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! An professor finds that the number of students enrolling in his class is approximated by s(t) = -tsquared + 20t + 80 where t is the number of tests given in one semester.
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Maximum:
You have a quadratic with a=-1,b=20,c=80
Max is at t=-b/2a = -20/-2 =10
# of students will be maximum if he gives 10 tests.
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No students means s = 0
-t^2+20t+80=0
t = [-20+-sqrt(400-4*-1*80]/(-2)
t = [-20+-sqrt(720)]/-2
t = [-20+-12sqrt5]/(-2)
Positive solution:
t = 23.42
Rounding down: No students if he gives 23 tests.
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Cheers,
Stan H.
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
An professor finds that the number of students enrolling
in his class is approximated by s(t) = -t² + 20t + 80
where t is the number of tests given in one semester.
Here is the graph. t, the number of tests is on the
x-axis and the number of students is on the y-axis.
(a) How many tests should the professor give so that the
number of students enrolling in the class is a maximum?
That will be the highest point, the vertex, of the graph.
The vertex is the point (h,k) where h =  ,
and k = s(h)
s(t) = -t² + 20t + 80
a = -1 (the coefficient of t²
b = 20 (the coefficient of t)
c = 80 (the constant)
h = =  = 10
k = s(h) = -10² + 20(10) + 80 = -100 + 200 + 80 = 180
The vertex (10,180) at the top of the graph is
interpreted as "when the professor gives 10 tests,
there will be 180 students.
(b) If the professor does not want students to enroll,
how many tests should he give?
We substitute 0 students for s(t) and solve for t:
s(t) = -t² + 20t + 80
0 = -t² + 20t + 80
t² - 20t - 80 = 0
That doesn't factor so we have to use the quadratic
formula:
______
-b ± Öb²-4ac
t = —————————————
2a
where a = 1; b = -20; c = -80
________________
-(-20) ± Ö(-20)²-4(1)(-80)
x = —————————————————————————————
2(1)
_______
20 ± Ö400+320
x = ————————————————
2
___
20 ± Ö720
x = ————————————
2
_____
20 ± Ö144·5
x = —————————————
2
_
20 ± 12Ö5
x = ———————————
2
_
20 12Ö7
x = ———— ± ——————
2 2
_
x = 10 ± 6Ö7
_
Using the +, x = 10 + 6Ö5, which
is one solution and equals about 23.41640786
_
Using the -, x = 10 - 6Ö5, which
is the other solution and equals about -3.416407865. This is
meaningless for this problem, since it's negative, so we
disregard it.
Therefore the point where the graph crosses the horizontal
axis is (23.4, 0) which means that if the professor gives
more than 23 tests, he will have no students.
Edwin
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