1. What are the dimensions of the box with largest volume which
can be made by cutting squares from the corners of an
8.5 "x 11" piece of cardboard and turning up the edges?

2. Find the dimensions of the rectangle of largest area that
has its base on the x axis and its other two vertices above
the x axis and lying on the curve f(x) = .

3. Which point on the graph of y = ex is closest to the origin?

4. 1. If you have 100 feet of fencing and you want to enclose a rectangular area
up against a long, straight wall, what is the largest area you can enclose?
y y
x
100 = 2y + x
A = xy
A = x(100 -x)
2
A(x) = 100x -x2
2
A'(x) = 50 - x
Max/Min Area when A'(x) = 0
i.e. when x = 50.
A'(x) + 0
0 50
Max area occurs when x = 50 and the maximum area is 50(50)/2 or 1250 sq feet

5. 2
2. A rectangular area of 3200 ft is to be fenced off. Two opposite sides will use
fencing costing $2 per foot and the remaining sides will use fencing costing$1 per
foot. Find the dimensions of least cost.
LW = 3200
W = 3200
W L
$1 / ft
L
$2 / ft
Cost = length x Cost per unit of length
Cost = 2L ( 2) + 2W (1)
C(L) = 4L + 2 3200
L
Max/Min cost will occur when C'(L) = 0
0
C'(L)
C'(L) = 4 - 6400
L2
0 40
C(L)
C'(L) = 0 = 4 - 6400
L2
When L = 40, W = 3200/40 = 80 so the cost is minimized if the
L2 = 1600 rectangular area has dimensions 40 ft x 80 ft.
L = 40

6. 3. A square-bottomed box with a top has a fixed volume of 500 cm 3.
What dimensions minimize the surface area?
V = 500 = x2h
h 500 = h
x2
Surface area S = 2x2 +4xh
x x = 2x2 +4x 500
x2
S = 2x2 + 2000
Minimize
x
S'(x) = 4x - 2000
x2
0 = 4x - 2000
x2
0 = 4(x3 - 500)
x2
S'(x) 0
x
= ∛500 = 7.94 cm
S(x) 7.94
when x = 7.94, the surface area is minimized
and h = 500/ 7.942 = 7.94
Dimensions of box which
minimizes surface area are
7.94 cm x 7.94 cm x 7.94 cm

7. 4. A square-bottomed box without a top has fixed volume
of 500 cm3. What dimensions minimize the surface area?
V = 500 = x2h
h 500 = h
x2
Surface area S = x2 +4xh
x x = x2 + 4x 500
x2
S = x2 + 2000
Minimize
x
S'(x) = 2x - 2000
x2
0 = 2x - 2000
x2 S'(x) 0
3
0 = 2(x - 1000)
x2 S(x) 10
x
= ∛1000 = 10 cm
when x = 10, the surface area is minimized
and h = 500/ 102 = 5
Dimensions of box which
minimizes surface area are
10 cm x 10 cm x 5 cm

8. (x, y)
(x, y) L
y
L
1-x (1,0)
L2 = (1- x)2 + y2
b
ut y = x2
so L2 = (1 - x)2 + x4
minimize this
2LdL = 2(1 - x)(-1) +
4x3
dL dx
dx =
(1 - x)(-1) +
dL 3
2x 0 = -1 + x + 2x3
dx
=
x = .589 (1-x)2 + x4 L'(x) 0
L(x) .589
Distance from (x,y) is a minimum when x =.589 and
y = .5892 = .347
So, the point on the parabola y = x2 which is closest
to the point (1, 0) is (.589, .347).

9. (x,y) L2 = (1 - x)2 + (y - 2)2
L = (1 - x)2 + ( 4 - x2 - 2)2
= (1 - x)2 + (2 - x2 )2
(1,2) N
eed to find the minimum
value of L
L2 = (1 - x)2 + (2 - x2)2
2
L L'(x) = 2(1 - x)(-1) + 2(2 - x2)(-2x)
L'(x) = -2 + 2x - 8x + 4x3
2 √ ((1 - x)2 + (2 - x2 )2)
L'(x) = -1 - 3x + 2x3
√ (1 - x)2 + (2 - x2 )2
L'(x) = 0 when 0 = -1 - 3x + 2x3
x = 1.366 if 0 < x < 2
Since L' (x) < 0 for x < 1.366 and L'(x) > 0 for
x > 1.366, L has a minimum value when x =
1.366.
When x = 1. 366, y = 4 - 1.3662 = 2.134
So (1.366, 2.134) is the point on the curve y = 4 - x2 which is
closest to the point (1,2)

10. SA = 210 = 2πr2 + 2πrh
h = 210 - 2πr2
h 2πr
r
V = πr2h
V(r) = πr2 210 - 2πr2
2πr
V(r) = r(210 - 2πr2)
Maximize
2
V'(r) = (210 - 6πr2)
2
V'(r) = 0 when 210 - 6πr2 = 0
210 = 6πr2 V'(r) 0
105 = r2
0 3.34
3π
√(35/π) = r = 3.34 cm
The volume is a maximum when r = 3.34 cm and
h = (210 - 2π(3.34)2 )/(2π(3.34)) = 6.68 cm

11. Minimum surface area when r = 5.42 cm and
h = 500/(π x 5.422) = 5.42 cm.

14. 11. A rectangle is bounded by the x and y axes and the line f(x) = 6 - x
2
What length and width should the rectangle have so that its area is a maximum?

15. 12. A right triangle is formed in the first quadrant by the x and y axes and a line
through the point (1,2). Find the vertices of the triangle so that its area is a minimum.

16. 13. Find the dimensions of the rectangle of largest area that has its base on the x axis
and its other two vertices above the x axis and lying on the parabola y = 8 - x2 .
(x, y)
y = 8 - x2

17. 14. A rectangle is drawn with sides parallel to the coordinate axes with its upper two
vertices on the parabola f(x) = 4 - x2 and its lower two vertices on the parabola g(x) = x2/2 -2.
What is the maximum possible area of the rectangle?
(x, f(x))
(x, g(x))

18. 15

19. 16.

20. HOMEWORK PROBLEMS

21. a) Amount of fencing available
Area constraints
2x2 ≥800 so x ≥ 20
y2 ≥100 so y ≥ 10
150 -1.5x ≥10
x≤ 140/1.5
x≤ 932/3
Domain of x
[20, 932/3]
Graph of A(x) has
A (932/3) = 17,522.2
A(20) = 15,200
no local max so
abs max value of A occurs
at an endpoint of
the domain. Max area enclosed is
17522. 2 ft2.