The J. R. Ryland Computer Company is considering a plant expansion to enable the company to begin production of a new computer product. The company’s president must determine whether to make the expansion a medium- or large-scale project. Demand for the new product is uncertain, which for planning purposes may be low demand, medium demand, or high demand. The probability estimates for demand are 0.20, 0.20, and 0.60, respectively. Letting x and y indicate the annual profit in thousands of dollars, the firm’s planners developed the following profit forecasts for the medium- and large-scale expansion projects. Medium-Scale Large-Scale
Expansion Profit Expansion Profit
x f(x) y f(y)
Low 50 0.2 0 0.2
Demand Medium 150 0.5 100 0.5
High 200 0.3 300 0.3
a. Compute the expected value for the profit associated with the two expansion alternatives.
Which decision is preferred for the objective of maximizing the expected profit?
b. Compute the variance for the profit associated with the two expansion alternatives.
Which decision is preferred for the objective of minimizing the risk or uncertainty?

Kindly check explanation

Step-by-step explanation:

Given the data:

Medium-Scale Large-Scale

Expansion Profit Expansion Profit

x f(x) y f(y)

Low 50 0.2 0 0.2

Demand Medium 150 0.5 100 0.5

High 200 0.3 300 0.3

a. Compute the expected value for the profit associated with the two expansion alternatives.

Which decision is preferred for the objective of maximizing the expected profit?

Expected value for medium scale expansion profit :

Expected value (E) = Σ(X) * f(x)

Σ[(50 * 0.2) + (150 * 0.5) + (200 * 0.3)]

= 145

Expected value for Large scale expansion profit :

Expected value (E) = Σ(X) * f(x)

Σ[(0 * 0.2) + (100 * 0.5) + (300 * 0.3)]

= 140

Medium scale expansion profit is preferred as it has the highest expected value.

b. Compute the variance for the profit associated with the two expansion alternatives.

Which decision is preferred for the objective of minimizing the risk or uncertainty?

Variance (V) = Σ(X - E)² * f(x):

Variance for medium scale expansion profit :

V = [((50-145)^2 * 0.2) + ((150-145)^2 * 0.5) + ((200-145)^2 * 0.3) = 2725

Variance for Large scale expansion profit :

V = [((0-140)^2 * 0.2) + ((100-140)^2 * 0.5) + ((300-140)^2 * 0.3) = 12400

Smaller variance is required to minimize risk, Hence, choose the medium scale expansion profit.

perimeter of the first rectangular board is 9 feet.

step-by-step explanation:

let p = perimeter of first board.

l = length of first board

w = width of first board

length of first board = 2 * width of first board - 3

l = 2w - 3

perimeter of first board = 2l + 2w

since l = 2w-3, then

p = 2*(2w-3) + 2w which equals 4w-6 + 2w which equals 6w-6

length and width of the second board are reciprocals of corresponding length and width of the first board.

length of second board = 1/(2w-3)

width of second board = 1/w

perimeter of second board is 1/5th perimeter of the first board.

perimeter of second board = p/5

perimeter of the second board = (2/(2w-3) + 2/w)

since the perimeter of the first board is 5 times the perimeter of the second board, this means that:

6w-6 = 5*(2/(2w-3) + 2/w)

which is the same as:

6w-6 = (10/(2w-3) + 10/w)

if we multiply both sides of this equation by (2w-3), we get:

(6w-6)*(2w-3) = 10 + 10*(2w-3)/w

if we multiply both sides of this equation by w, we get:

(6w-6)*(2w-3)*w = 10*w + 10*(2w-3)

we can simplify this to become:

(6w-6)*(2w-3)*w = 10w + 20w - 30

which becomes:

(6w-6)*(2w-3)*w = 30w-30

if we divide both sides of this equation by (6w-6), we get:

(2w-3)*w = (30w-30)/(6w-6)

which becomes:

(2w-3)*w = 5

this can be simplified to:

2w^2 - 3w = 5

subtract 5 from both sides of this equation to get:

2w^2 - 3w - 5 = 0

which can be factored into:

(2w-5)*(w+1) = 0

which makes:

w = 5/2

or

w = -1.

w can't be negative so the only possible answer is w = 5/2.

if w = 5/2, then

p = 2w-3 = 5-3 = 2

we have:

p = 2

w = 5/2

p = 2l + 2w gets p = 4 + 5 = 9

the perimeter of the first board is 9 feet.

since the perimeter of the second board is 1/5 the perimeter of the first board, the perimeter of the second board is 9/5.

to see if that's correct, we substitute known values for l and w into the second board.

the length of the second board is 1/l.

this becomes 1/2.

the width of the second board is 1/w

this bgecomes 1/(5/2) = (2/5)

let l2 = length of second board.

let w2 = width of second board.

l2 = (1/2)

w2 = (2/5)

let p2= perimeter of second board.

then p2 = 9/5

since p2 = 2*l2 + 2*w2, we should get p2 = 9/5 using the dimensions of l2 and w2.

2 * l2 = 2 * (1/2) = 1

2 * w2 = 2 * (2/5) = 4/5

1 is the same as 5/5

5/5 + 4/5 = 9/5

p2 checks out good.

the dimensions as given are good.

answer to the problem is:

perimeter of the first rectangular board is 9 feet.

a

step-by-step explanation: