1. Extra Quadratic Min/Max
Problems and Solutions

2. A cattle farmer wants to build a rectangular
fenced enclosure divided into five rectangular
pens as shown in the diagram. A total length
of 120m of fencing material is available. Find
the overall dimensions of the enclosure that
will make the total area a maximum.

3. Solution L
w w w w w w
L
120 = 6w + 2L L = 120 - 6w
Area = w*L 2
L = (60 - 3w)
Area = w(60-3w)
Area = 60w - 3w 2

4. 2
Area = 60w - 3w
Area = -3{w 2 - 20w}
Area = -3{w 2 - 20w + 100 -100}
2 - 20w + 100) -100}
Area = -3{(w
2 - 100}
Area = -3{(w - 10)
Area = -3(w - 10) 2 - (-3)100
2
Area = -3(w - 10) +300

5. L
w w w w w w
L
2
Area = -3(w - 10) +300
Area is maximum when w = 10m
The maximum area is 300m 2
Area = L*w
300 = L*10
L = 30m

6. An amusement park charges $8 admission
and averages 2000 visitors per day. A
survey shows that, for each $1 increase in
the admission cost, 100 fewer people will
visit the park.
What admission cost gives the maximum
profit?

7. Solution
Profit = Admission * Visitors
A = 8 + 1x
V = 2000 - 100x
where x = number of times the price is
increased
P = (8+x)(2000-100x)

8. P = (8+x)(2000-100x)
P = 16000 - 800x + 2000x - 100x2
P = -100x 2 + 1200x + 16000
P = -100{x2 - 12x} + 16000
2 - 12x + 36) -36} +16000
P = -100{(x

9. P = -100{(x 2 - 12x + 36) -36} +16000
P = -100{(x-6)2 -36} + 16000
P = -100(x-6)2 - (-100)36 + 16000
2 + 19600
P = -100(x-6)

10. Profit = Admission * Visitors
A = 8 + 1x
V = 2000 - 100x
where x = number of times the price is
increased
P = -100(x-6) 2 + 19600
So maximum profit is $19600
when x = 6
The admission price should be
A = 8 + 6 = $14

11. Determine the maximum area of a
triangle if the sum of its base and its
height is 13 cm.

12. Solution
b + h = 13 b = (13 - h)
Area = 1/2 b*h
Area = 1/2(13-h)(h)
Area = (6.5 - .5h)(h)
Area = 6.5h - .5h2

13. 2
Area = 6.5h - .5h
Area = -.5b 2 + 6.5h
Area = -.5{h2 - 13h}
Area = -.5{(h2 - 13h + 6.52) - 6.52}
Area = -.5{(h-6.5) 2 - 6.52}
2 - (-.5)6.52
Area = -.5(h-6.5)

14. 2 - (-.5)6.52
Area = -.5(h-6.5)
2 + 21.125
Area = -.5(h-6.5)
The maximum area of the triangle is
21.125 cm2 when the base is 6.5 cm.