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

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) = .

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

x

yy

100 = 2y + x

A = xyA = x(100 -x) 2A(x) = 100x -x2

2A'(x) = 50 - x

Max/Min Area when A'(x) = 0 i.e. when x = 50.

0 50

A'(x) + 0

Max area occurs when x = 50 and the maximum area is 50(50)/2 or 1250 sq feet

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?

2. A rectangular area of 3200 ft2 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.

L

W

LW = 3200 W = 3200 L

$2 / ft

$1 / 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

C'(L) = 4 - 6400 L2

C'(L) = 0 = 4 - 6400 L2

L2 = 1600L = 40

0 40

C'(L)0

C(L)

When L = 40, W = 3200/40 = 80 so the cost is minimized if the rectangular area has dimensions 40 ft x 80 ft.

3. A square-bottomed box with a top has a fixed volume of 500 cm3. What dimensions minimize the surface area?

x x

hV = 500 = x2h 500 = h x2

Surface area S = 2x2 +4xh = 2x2 +4x 500 x2

S = 2x2 + 2000 x

Minimize

S'(x) = 4x - 2000 x2

0 = 4x - 2000 x2

0 = 4(x3 - 500) x2

x = ∛500 = 7.94 cm

7.94

S'(x) 0

S(x)

when x = 7.94, the surface area is minimized and h = 500/ 7.942 = 7.94

Dimensions of box which minimizes surface area are7.94 cm x 7.94 cm x 7.94 cm

4. A square-bottomed box without a top has fixed volume of 500 cm3. What dimensions minimize the surface area?

x x

hV = 500 = x2h 500 = h x2

Surface area S = x2 +4xh = x2 + 4x 500 x2

S = x2 + 2000 x

Minimize

S'(x) = 2x - 2000 x2

0 = 2x - 2000 x20 = 2(x3 - 1000)

x2

x = ∛1000 = 10 cm10

S'(x) 0

S(x)

when x = 10, the surface area is minimized and h = 500/ 102 = 5

Dimensions of box which minimizes surface area are10 cm x 10 cm x 5 cm

(1,0)

L

(x, y) L

(x, y)

y

1-x

L2 = (1- x)2 + y2

but y = x2

so L2 = (1 - x)2 + x4

minimize this

2LdL = 2(1 - x)(-1) + 4x3

dx (1 - x)(-1) + 2x3

(1-x)2 + x4

dLdx =

dLdx =0 = -1 + x + 2x3

x = .589

.589

L'(x) 0

L(x)

Distance from (x,y) is a minimum when x =.589 andy = .5892 = .347

So, the point on the parabola y = x2 which is closestto the point (1, 0) is (.589, .347).

(1,2)

(x,y)

LL2 = (1 - x)2 + (y - 2)2

= (1 - x)2 + ( 4 - x2 - 2)2

= (1 - x)2 + (2 - x2 )2

Need to find the minimum value of L

L2 = (1 - x)2 + (2 - x2)2

2L 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)

h

r

SA = 210 = 2πr2 + 2πrh h = 210 - 2πr2

2πr

V = πr2hV(r) = πr2 210 - 2πr2

2πrV(r) = r(210 - 2πr2) 2 Maximize

V'(r) = (210 - 6πr2) 2V'(r) = 0 when 210 - 6πr2 = 0 210 = 6πr2

105 = r2

3π √(35/π) = r = 3.34 cm

3.34

V'(r)

0

0

The volume is a maximum when r = 3.34 cm and h = (210 - 2π(3.34)2 )/(2π(3.34)) = 6.68 cm

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

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?

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.

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

14. A rectangle is drawn with sides parallel to the coordinate axes with its upper twovertices 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))

15

16.

HOMEWORK PROBLEMS

2x2 ≥800 so x ≥ 20 y2 ≥100 so y ≥ 10 150 -1.5x ≥10 x≤ 140/1.5 x≤ 932/3

A (932/3) = 17,522.2A(20) = 15,200

Max area enclosed is

17522. 2 ft2.

Graph of A(x) has

no local max so

abs max value of A occurs

at an endpoint of

the domain.

a) Amount of fencing available

Area constraints

Domain of x [20, 932/3]