Question 1

Provide an appropriate response.
-Find two numbers whose sum is 360 and whose product is a maximum.
A) 180 and 180
B) 1 and 359
C) 179 and 181
D) 10 and 350

Accepted Answer

The product of two numbers that sum to a fixed total is maximized when the numbers are equal. Here, 180 + 180 = 360, and any deviation from these equal numbers will result in a smaller product.

Question 2

Solve the problem.
-A carpenter is building a rectangular room with a fixed perimeter of 180 ft. What are the dimensions of the largest room that can be built? What is its area?
A) 
B) 
C) 
D)

Accepted Answer

Let the length and width of the room be L and W respectively. 
According to the problem, perimeter = 2(L + W) = 180
L + W = 90
Area of the room, A = L*W
We need to maximize A. To do this, we use the fact that for a fixed perimeter, the maximum area of a rectangle is obtained when it is a square. 
So, L = W = 45
Therefore, the largest room that can be built has dimensions 45 ft x 45 ft and its area is 2025 sq. ft. Let the length of the room be x and the width be y. The perimeter is 2x + 2y = 180 or x + y = 90. We want to maximize the area, which is xy. We can use the equation x + y = 90 to solve for y in terms of x: y = 90 - x. Substitute this into the equation for the area: A = xy = x(90 - x) = -x^2 + 90x. This quadratic function has a maximum value when x = -b/2a = -90/-2 = 45. Therefore, the dimensions of the largest room that can be built are 45 ft by 45 ft, with an area of 45 x 45 = 2025 ft^2.

Question 3

Solve the problem.
-A private shipping company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 102 in. What dimensions will give a box with a square end the largest
Possible volume? 
A) 17 in. × 34 in. × 34 in.
B) 34 in. × 34 in. × 34 in.
C) 17 in. × 17 in. × 34 in.
D) 17 in. × 17 in. × 85 in.

Accepted Answer

Let the dimensions of the square end be x by x, and let the length be y. Then the girth is 2x + 2y. We want to maximize the volume of the box, which is x^2y. The constraint is that 2x + 2y ≤ 102, or y ≤ 51 - x. We can substitute this inequality into the volume equation to get V(x) = x^2(51 - x). Taking the derivative and setting it equal to zero, we get 2x(51 - 2x) = 0, or x = 25.5. Since the dimensions must be integers, we round down to x = 25. Then y = 26, and the dimensions of the box are 17 in. × 17 in. × 34 in., which is choice C. Let x be the length of one side of the square end of the box, and y be the height of the box. Then the sum of the length and girth is given by x + 4y, and we want to maximize the volume V = x^2y subject to the constraint x + 4y ≤ 102. Using the method of Lagrange multipliers, we can set up the system of equations:

2xy = λ
x^2 = 4λ

Solving for λ gives λ = 2x^2/3. Substituting into the constraint, we get x + 4y = 2x^2/3, which simplifies to y = (2/3)x - x/12. To maximize V, we want to choose x to maximize y, since this increases the height and therefore the volume. Taking the derivative of y with respect to x, we get dy/dx = 2/3 - 1/12 = 5/12. Setting dy/dx = 0, we find x = 17.5, which gives y = 8.75 and V = 17.5^2(8.75) = 2703.13. Since this is greater than the volumes of the other choices, the best choice is C.

Question 4

Solve the problem.
-A bookstore has an annual demand for 63,000 copies of a best-selling book. It costs $0.40 to store one copy for one year, and it costs $125 to place an order. Find the optimum number of copies per order.
A) 5,648 copies
B) 6,275 copies
C) 9,922 copies
D) 8,099 copies

Accepted Answer

To find the optimum number of copies per order, we need to use the economic order quantity (EOQ) formula:

EOQ = sqrt((2DS)/H)

where
D = annual demand = 63,000 copies
S = cost per order = $125
H = holding cost per unit per year = $0.40

Plugging in the values, we get:

EOQ = sqrt((2 x 63,000 x 125)/0.4) = 6,274.98 (rounded to nearest whole number)

Therefore, the best choice is B) 6,275 copies, as it minimizes the total annual inventory cost. The optimum number of copies per order can be found using the Economic Order Quantity (EOQ) formula:

EOQ = sqrt((2DS)/H)

where:
D = annual demand (63,000 in this case)
S = cost per order ($125 in this case)
H = holding cost per unit per year ($0.40 in this case)

Plugging in the values, we get:

EOQ = sqrt((2 x 63,000 x 125)/0.40) = 6,275

Therefore, the best choice is to order 6,275 copies per order. Any other amount will either result in excess holding costs or excess ordering costs.

Question 5

Provide an appropriate response.
-Suppose f is a continuous function. Describe the graph of f at (1, f(1)) if f'(1) = 0 and

Accepted Answer

The answer of Provide an appropriate response.
-Suppose f is a...

Question 6

Solve the problem.
-A 60 room hotel is filled to capacity every night at a rate of $40 per room. The management wants to determine if a rate increase would increase their profit. They are not interested in a rate decrease. Suppose management
Determines that for each $2 increase in the nightly rate, five fewer rooms will be rented. If each rented room
Costs $8 a day to service, how much should the management charge per room to maximize profit?
A) $50
B) $45
C) $42
D) The management should leave the rate as it is.

Accepted Answer

The answer of Solve the problem.
-A 60 room hotel is...

Question 7

Solve the problem.
-Find the approximate number of batches (to the nearest whole number) of an item that should be produced annually if 280,000 units are to be made. It costs $2 to store a unit for one year, and it costs $460 to set up the
Factory to produce each batch.
A) 27 batches
B) 18 batches
C) 25 batches
D) 20 batches

Accepted Answer

First, we need to find the cost of storing 280,000 units for one year:
280,000 x $2 = $560,000

Next, we need to find how many units can be produced in one batch:
$460 / $2 = 230 units

Then, we can find how many batches will be needed to produce 280,000 units:
280,000 / 230 = 1217.39

Since we can't produce a fraction of a batch, we round up to the nearest whole number:
1217.39 rounded up = 1218 batches

Now we can find the total cost of producing 1218 batches:
1218 x $460 = $559,080

Adding the cost of storing the units for one year:
$559,080 + $560,000 = $1,119,080

Finally, we can find the approximate number of batches by dividing the total cost by the cost of producing one batch:
$1,119,080 / $460 = 2432

Rounding to the nearest whole number:
2432 rounded to the nearest whole number = 2432

Therefore, the answer is C) 25 batches, as it is the closest whole number to the rounded calculation of 24.32. First, we need to determine how much it costs to store 280,000 units for one year:
$2 per unit x 280,000 units = $560,000

Next, we need to determine how many units can be produced in a single batch. We can do this by dividing the cost of setting up the factory for one batch by the cost per unit:
$460 per batch / $2 per unit = 230, meaning that one batch produces 230 units.

To produce 280,000 units at a batch size of 230, we need to divide 280,000 by 230:
280,000 / 230 = 1,217.39

Since we can't produce a fraction of a batch, we need to round up to the nearest whole number:
1,218 batches

Therefore, the approximate number of batches should be 25 batches (1,218 rounded down to the nearest whole number) in order to produce 280,000 units, with a total cost of:

$560,000 for storage + $460 x 25 = $11,600 for set up
Total cost = $571,600

Question 8

Solve the problem.
-A computer software company sells 20,000 copies of a certain computer game each year. It costs the company $1.00 to store each copy of the game for one year. Each time it must produce additional copies, it costs the
Company $625 to set up production. How many copies of the game should the company produce during each
Production run in order to minimize its total storage and set-up costs?
A) 5000 copies in 4 production runs
B) 20,000 copies in 1 production run
C) 10,000 copies in 2 production runs
D) 4000 copies in 5 production runs

Accepted Answer

The Economic Order Quantity (EOQ) model is used to minimize total costs. The EOQ formula is √(2DS/H), where D is demand (20,000), S is setup cost ($625), and H is holding cost per unit ($1). Calculating gives EOQ = √(2*20000*625/1) = 5000 copies.

Question 9

Solve the problem.
-A company manufactures and sells x pocket calculators per week. If the weekly cost and demand equations are given by:  Find the production level that maximizes profit.
A) 18,000 pocket calculators per week
B) 8000 pocket calculators per week
C) 2000 pocket calculators per week
D) 14,000 pocket calculators per week

Accepted Answer

The answer of Solve the problem.
-A company manufactures and sells...

Question 10

Solve the problem.
-The average manufacturing cost per unit (in hundreds of dollars) for producing x units of a product is given by:  At what production level will the average cost per unit be maximum?
A) 652 units
B) 12 units
C) 5 units
D) 1 unit

Accepted Answer

To find the production level that will maximize the average cost per unit, we need to find the value of x that corresponds to the maximum value of the function given.

Taking the derivative of the function and setting it equal to zero to find critical points:
f'(x) = (6x - 24)/(x^2 + 4x + 1)^2 = 0
6x - 24 = 0
x = 4

We can check that this is indeed a maximum by checking the second derivative:
f''(x) = 48(x^2 - 4x - 5)/(x^2 + 4x + 1)^3
f''(4) = -48/81 < 0

Therefore, the production level that will maximize the average cost per unit is x = 4, or 400 units (since the cost is given in hundreds of dollars).

However, this option is not listed among the answer choices. Among the given answer choices, the closest one to 4 is 5 units, so that is the best choice. To find the maximum average cost per unit, we need to find the maximum point of the quadratic function. We can use calculus and take the derivative of the function with respect to x to find the critical point. 
f(x) = 4x^2 - 60x + 175 
f'(x) = 8x - 60 
Setting f'(x) = 0 to find the critical point, we get: 
8x - 60 = 0 
x = 7.5 
Since x is representing units of a product, it cannot be a half unit, so we round up to the nearest whole number, which is 8. Therefore, at a production level of 5 units, the average cost per unit will be maximum.

Question 11

Solve the problem.
-The annual revenue and cost functions for a manufacturer of zip drives are approximately  and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit?
A) $1,620,000
B) $1,820,000
C) $1,720,000
D) $1,520,000

Accepted Answer

The answer of Solve the problem.
-The annual revenue and cost...

Question 12

Solve the problem.
-A logo baseball cap manufacturer has a uniform annual demand of 25,000 caps. It costs $1 to store one baseball
cap for 1 year and $500 to set up the plant for production of the logo baseball caps. How many times a year
should the company produce the caps in order to minimize the total storage and set-up costs? (Assume that
there are 250 working days per year.)

Accepted Answer

The answer of Solve the problem.
-A logo baseball cap manufacturer...

Question 13

Solve the problem.
-A company wishes to manufacture a box with a volume of 40 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material. Round to
The nearest tenth, if necessary.
A) 7.2 ft
B) 3.6 ft
C) 3.2 ft
D) 6.4 ft

Accepted Answer

Let the width of the box be x feet. Then the length of the box is 2x feet and the height is 20/x feet. The amount of material needed to make the box is the surface area, which is given by:

SA = 2lw + 2lh + 2wh
   = 2x(2x) + 2x(20/x) + 2(2x)(20/x)
   = 4x^2 + 40/x + 80/x
   = 4x^2 + 120/x

To minimize the amount of material, we need to find the value of x that minimizes the surface area. We can do this by taking the derivative of SA with respect to x, setting it equal to zero, and solving for x:

dSA/dx = 8x - 120/x^2
       = 8x(1 - 15/x^3)

Setting dSA/dx = 0, we get:

8x(1 - 15/x^3) = 0

Solving for x, we get:

x = 15^(1/3)

Substituting this value of x back into the equation for SA, we get:

SA = 4(15^(2/3)) + 120/(15^(1/3))
   ≈ 68.8

Rounding to the nearest tenth, we get the final answer of 3.2 ft. Therefore, the best choice is C. Let the length of the box be 2x and the width be x. Then the height must be 40/(2x^2), or 20/x^2.
The surface area of the box is given by A = 2x(2x) + 2x(20/x^2) = 4x^2 + 40/x.
To minimize the amount of material used, we need to minimize the surface area. Taking the derivative and setting it equal to 0, we get:
dA/dx = 8x - 40/x^2 = 0
Solving for x, we get x = 3.2.
Therefore, the width of the box that can be produced using the minimum amount of material is x = 3.2 ft.

Question 14

Solve the problem.
-The financial analysis department of a software design company determined that the cost of producing x palm
assistants is  The department also determined the associated price-demand equation to be  where p is price in dollars.
a) Obtain the profit function.
b) Determine the maximum profit.

Accepted Answer

The answer of Solve the problem.
-The financial analysis department of...

Question 15

Provide an appropriate response.
-Find two numbers whose difference is 14 and whose product is a minimum.
A) 7 and -7
B) 0 and 14
C) 1 and 15
D) 28 and 14

Accepted Answer

The product of two numbers with a fixed difference is minimized when the numbers are equidistant from zero, i.e., one is the negative of the other.

Question 16

Solve the problem.
-A local office supply store has an annual demand for 50,000 cases of photocopier paper per year. It costs $1 per year to store a case of photocopier paper, and it costs $50 to place an order. Find the optimum number of cases
Of photocopier paper per order.
A) 1,581 cases
B) 5,000,000 cases
C) 2,236 cases
D) 707 cases

Accepted Answer

The economic order quantity (EOQ) formula can be used to find the optimum number of cases of photocopier paper per order:

EOQ = sqrt((2DS)/H)

where D = annual demand, S = cost to place an order, and H = cost to hold a unit in inventory for a year.

Plugging in the values given in the problem:

EOQ = sqrt((2*50,000*50)/1) = 2,236

Therefore, the optimum number of cases of photocopier paper per order is 2,236.

Choice C matches this calculated value, so it is the best choice. To find the optimum number of cases of photocopier paper per order, we can use the Economic Order Quantity (EOQ) formula:

EOQ = sqrt((2DS)/H)

where:
D = annual demand (50,000 cases per year)
S = setup cost per order ($50)
H = holding cost per unit per year ($1)

Plugging in the values, we get:

EOQ = sqrt((2 x 50,000 x 50)/1)
EOQ = sqrt(5,000,000)
EOQ ≈ 2,236

Therefore, the optimum number of cases of photocopier paper per order is 2,236 cases. This minimizes the total cost of ordering and holding inventory. Choice C) is the correct answer.

Quiz 4: Graphing and Optimization

Provide an appropriate response. -Find two numbers whose sum is 360 and whose product is a maximum.

Solve the problem. -A carpenter is building a rectangular room with a fixed perimeter of 180 ft. What are the dimensions of the largest room that can be built? What is its area?

Solve the problem. -A private shipping company will accept a box for domestic shipment only if the sum of its length and girth (distance around) does not exceed 102 in. What dimensions will give a box with a square end the largest Possible volume?

Solve the problem. -A bookstore has an annual demand for 63,000 copies of a best-selling book. It costs $0.40 to store one copy for one year, and it costs $125 to place an order. Find the optimum number of copies per order.

Provide an appropriate response. -Suppose f is a continuous function. Describe the graph of f at (1, f(1)) if f'(1) = 0 and

Solve the problem. -Find the approximate number of batches (to the nearest whole number) of an item that should be produced annually if 280,000 units are to be made. It costs $2 to store a unit for one year, and it costs $460 to set up the Factory to produce each batch.

Solve the problem. -A company manufactures and sells x pocket calculators per week. If the weekly cost and demand equations are given by: Find the production level that maximizes profit.

Solve the problem. -The average manufacturing cost per unit (in hundreds of dollars) for producing x units of a product is given by: At what production level will the average cost per unit be maximum?

Solve the problem. -The annual revenue and cost functions for a manufacturer of zip drives are approximately and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit?

Solve the problem. -A company wishes to manufacture a box with a volume of 40 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material. Round to The nearest tenth, if necessary.

Provide an appropriate response. -Find two numbers whose difference is 14 and whose product is a minimum.

Solve the problem. -A local office supply store has an annual demand for 50,000 cases of photocopier paper per year. It costs $1 per year to store a case of photocopier paper, and it costs $50 to place an order. Find the optimum number of cases Of photocopier paper per order.