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Mathematics : asked on tonya3498
 09.07.2022

what is the minimum unit cost? do not round your answer

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Mathematics
an aircraft factory manufactures airplane engines. the unit cost c (the cost in dollars to make each airplane engine) depends on the number of engines made. if x engines are made, then the unit cost is given by the function c(x) = 1.1x^2 - 638x + 111,541 . what is the minimum unit cost? do not round your answer.
Step-by-step answer
P Answered by PhD

=65  181/290

Step-by-step explanation:

The minimum cost is arrived to when ΔC/Δx is zero.

ΔC/Δx= 2.2x-638

We equate this to zero to get the minimum.

2.2x-638=0

2.2x=638

x=290

The cost for 290 engines is given by:

C(x)=1.1x²-638x+111,541

We replace the value of x above in this function.

=1.1(290)²-638(290)+111541

=19 031

Then 1 unit costs 19031/290

=65  181/290

Mathematics
A medical equipment industry manufactures X-ray machines. The unit cost (the cost in dollars to make each X-ray machine) depends on the number of machines made. If machines are made, then the unit cost is given by the function . What is the minimum unit cost?Do not round your answer.
Step-by-step answer
P Answered by PhD

$14,362

Step-by-step explanation:

The computation of the minimum unit cost is shown below:

Given that

0.6x^2 - 108x + 19,222

And as we know that the quadratic equation form is

ax^2 + bx + c

where

a = 0.6

b = -108

c = 19,222

Now for determining the minimal cost we applied the following formula which is

= \frac{-b}{2a} \\\\ = \frac{-(-108)}{2\times 0.6} \\\\ = \frac{108}{1.2}

= 90

Now put these values to the above equation

= 0.6\times 90^{2} - 108 \times 90 + 19,222

= 14,362

Mathematics
Amedical equipment industry manufactures x-ray machines. the unit cost c (the cost in dollars to make each x-ray machine) depends on the number of machines made. if x machines are made, then the unit cost is given by the function c(x)=0.6x^2-108x+19,222. what is the minimum unit cost? do not round your answer
Step-by-step answer
P Answered by PhD

Minimum Unit Cost = $14,362

Step-by-step explanation:

The standard form of a quadratic is given by:

ax^2 + bx + c

So for our function, we can say,

a = 0.6

b = -108

c = 19,222

We can find the vertex (x-coordinate where minimum value occurs) by the formula -b/2a

So,

-(-108)/2(0.6) = 108/1.2 = 90

Plugging this value into original function would give us the minimum (unit cost):

c(x)=0.6x^2-108x+19,222\\c(90)=0.6(90)^2-108(90)+19,222\\=14,362

Mathematics
an aircraft factory manufactures airplane engines. the unit cost c (the cost in dollars to make each airplane engine) depends on the number of engines made. if x engines are made, then the unit cost is given by the function c(x) = 1.1x^2 - 638x + 111,541 . what is the minimum unit cost? do not round your answer.
Step-by-step answer
P Answered by PhD

=65  181/290

Step-by-step explanation:

The minimum cost is arrived to when ΔC/Δx is zero.

ΔC/Δx= 2.2x-638

We equate this to zero to get the minimum.

2.2x-638=0

2.2x=638

x=290

The cost for 290 engines is given by:

C(x)=1.1x²-638x+111,541

We replace the value of x above in this function.

=1.1(290)²-638(290)+111541

=19 031

Then 1 unit costs 19031/290

=65  181/290

Mathematics
A vehicle factory manufactures cars. The unit cost (the cost in dollars to make each car) depends on the number of cars made. If cars are made, then the unit cost is given by the function . What is the minimum unit cost? Do not round your answer.
Step-by-step answer
P Answered by Specialist

The function is missing in the question. The function is $ C(x) = 0.8x^2 -544x +97410 $

The answer is 4930

Step-by-step explanation:

Unit cost of a car is given as

$ C(x) = 0.8x^2 -544x +97410 $

Cost will be minimum when

x = -(-544)/ 2 x 0.8

  = 340

Therefore, minimum cost for unit car is

$ C(x) = 0.8x^2 -544x +97410 $

$ = 0.8(340)^2 -544(340) +97410 $

=  4930

Mathematics
A medical equipment industry manufactures X-ray machines. The unit cost c (the cost in dollars to make each X-ray machine) depends on the number of machines made. If x machines are made, then the unit cost is given by the function c(x)=x^2-520x+72857. What is the minimum unit cost? Do not round your answer
Step-by-step answer
P Answered by PhD

The minimum unit cost is $5257

Step-by-step explanation:

Minimization

Given a function c(x), the minimum value of c can be found by computing the first derivative. Equating the first derivative to zero will provide the critical points, or candidate point to maximize or minimize the function. The second derivative criterion will make clear which type of point was obtained.

The cost in dollars to produce x machines is

C(x)=x^2-520x+72857

Find the first derivative

C'(x)=2x-520

Equate to 0

2x-520=0

Solving:

x=260

There must be produced 260 machines to minimize the cost. The minimum cost is

C(260)=260^2-520\cdot 260+72857

C(260)=5257

The minimum unit cost is $5257

Mathematics
Asap a vehicle factory manufactures cars. the unit cost c (the cost in dollars to make each car) depends on the number of cars made. if x cars are made, then the unit cost is given by the function c(x) =x^2-320x+40,052 . what is the minimum unit cost? do not round your answer.
Step-by-step answer
P Answered by PhD
The C(x) equation is a parabola with a squared "x" and a positive leading term coefficient, meaning, is a vertical parabola and it opens upwards

the lowest point, or lowest cost value, it reaches, is the U-turn point, or vertex

\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{llccll}
y = &{{ 1}}x^2&{{ -320}}x&{{ +40052}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad 
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)

so, the lowest cost is at   \bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}
Mathematics
Avehicle factory manufactures cars. the unit cost c (the cost in dollars to make each car) depends on the number of cars made. if x cars are made, then the unit cost is given by the function c(x) = x^2 - 400x + 45,377 . what is the minimum unit cost? do not round your answer.
Step-by-step answer
P Answered by PhD

The minimum unit cost is 5377

Step-by-step explanation:

Note that we have a cudratic function of negative principal coefficient.

The minimum value reached by this function is found in its vertex.

For a quadratic function of the form

ax ^ 2 + bx + c

the x coordinate of the vertex is given by the following expression

x=-\frac{b}{2a}

In this case the function is:

C(x) = x^2 - 400x + 45,377

So:

a=1\\b=-400\\c=45,377

Then the x coordinate of the vertex is:

x=-\frac{-400}{2(1)}

x=200\ cars

So the minimum unit cost is:

C(200) = (200)^2 - 400(200) + 45,377

C(200) = 5377

Business
Bell computers purchases integrated chips at ​$350 per chip. the holding cost is ​$35 per unit per​ year, the ordering cost is ​$119 per​ order, and sales are steady at 405 per month. the​ company s supplier, rich blue chip​ manufacturing, inc., decides to offer price concessions in order to attract larger orders. the price structure is shown below. rich blue​ chip s price structure quantity purchased ​ price/unit ​ 1-99 units ​$350 ​ 100-199 units ​$325 200 or more units ​$300 ​a) what is the most​ cost-effective
Step-by-step answer
P Answered by Master
When holding costs are $35 per unit,

1.The most cost-effective order quantity (assuming they take the most cost-effective discount, and use a fixed holding cost) is 182 units.

2. At the chosen level of quantity, discount, and using the fixed holding cost, the total annual cost for Bell computers to order, purchase, and hold the integrated chips is $1,585,898.

When holding costs are 10% of purchase price per unit,

1.The most cost-effective order quantity (assuming they take the most cost-effective discount, and use a fixed holding cost) is 189 units.

2. At the chosen level of quantity, discount, and using the fixed holding cost, the total annual cost for Bell computers to order, purchase, and hold the integrated chips is $1,585,546.

When holding costs are $35 per unit,

We follow these steps to arrive at the

1. We have

Ordering Costs per order  $119

Holding Cost per unit            $35

Demand per month            405 units

Demand per year is                                405*12 = 4860 units

Since the most cost-effective order quantity is the Economic Order Quantity (EOQ), we compute the EOQ

\mathbf{EOQ = \sqrt{\frac{2SD}{H}}}

where

D is  demand per year

S is the Ordering cost per order

H is the holding cost per unit

Substituting the values we get,

EOQ = \sqrt{\frac{2*(405*12)*119}{35}}

EOQ = \sqrt{33048}

\mathbf{EOQ = 181.7910889 \approx 182 units}

2. The annual costs of ordering, purchasing and holding the integrated chips is the sum of the cost of ordering, purchasing and holding the integrated  chips.

Since the EOQ at 182 units falls in the second slab of Rich Blue Manufacturing, Bell computer can purchase chips at $325 per unit

Cost of purchasing the chips     \mathbf{405*12*325 = 1579500}

Number of orders to placed         \frac{Annual Demand}{EOQ}

Number of orders to placed         \frac{405*12}{182}

Number of orders to placed   \mathbf{26.703 \approx 27 orders}

Cost of orders                           Number of orders * Cost per order

Cost of orders                               \mathbf{27 * 119 = 3213
}

Holding Costs                        \frac{EOQ}{2} * Holding cost per unit

Holding Costs                               \frac{182}{2} * 35

Holding Costs                               \mathbf{3185
}

Total annual costs       \mathbf{1579500 + 3213 + 3185 = 1585898}

When holding costs are 10% of purchase price per unit,

1. We need to calculate the EOQ, which holding cost at each purchase price

EOQ_{350} = (2*(405*12)*119)/(350*0.10) \approx 182 units

\mathbf{EOQ_{325} = (2*(405*12)*119)/(325*0.10)\approx 189 units}

EOQ_{300} = (2*(405*12)*119)/(300*0.10)\approx 197 units

Since the EOQ lies between 100 and 199 units in all the three costs, Bell Computers can purchase the units only at $325 per unit, so its holding cost will be 10% of $325, which is $32.50 per unit.

2.2. The annual costs of ordering, purchasing and holding the integrated chips is the sum of the cost of ordering, purchasing and holding the integrated  chips.

Since the EOQ at 189 units falls in the second slab of Rich Blue Manufacturing, Bell computer can purchase chips at $325 per unit

Cost of purchasing the chips     \mathbf{405*12*325 = 1579500}

Number of orders to placed         \frac{Annual Demand}{EOQ}

Number of orders to placed         \frac{405*12}{189}

Number of orders to placed   \mathbf{24.762 \approx 25 orders}

Cost of orders                           Number of orders * Cost per order

Cost of orders                               \mathbf{25 * 119 = 2975 }

Holding Costs                        \frac{EOQ}{2} * Holding cost per unit

Holding Costs                               \frac{189}{2} * 32.5

Holding Costs                               \mathbf{3071.25
}

Total annual costs    \mathbf{1579500 + 2975 + 3071.25 = 1585546.25
}

Business
Bell computers purchases integrated chips at ​$350 per chip. the holding cost is ​$35 per unit per​ year, the ordering cost is ​$119 per​ order, and sales are steady at 405 per month. the​ company s supplier, rich blue chip​ manufacturing, inc., decides to offer price concessions in order to attract larger orders. the price structure is shown below. rich blue​ chip s price structure quantity purchased ​ price/unit ​ 1-99 units ​$350 ​ 100-199 units ​$325 200 or more units ​$300 ​a) what is the most​ cost-effective
Step-by-step answer
P Answered by Specialist
When holding costs are $35 per unit,

1.The most cost-effective order quantity (assuming they take the most cost-effective discount, and use a fixed holding cost) is 182 units.

2. At the chosen level of quantity, discount, and using the fixed holding cost, the total annual cost for Bell computers to order, purchase, and hold the integrated chips is $1,585,898.

When holding costs are 10% of purchase price per unit,

1.The most cost-effective order quantity (assuming they take the most cost-effective discount, and use a fixed holding cost) is 189 units.

2. At the chosen level of quantity, discount, and using the fixed holding cost, the total annual cost for Bell computers to order, purchase, and hold the integrated chips is $1,585,546.

When holding costs are $35 per unit,

We follow these steps to arrive at the

1. We have

Ordering Costs per order  $119

Holding Cost per unit            $35

Demand per month            405 units

Demand per year is                                405*12 = 4860 units

Since the most cost-effective order quantity is the Economic Order Quantity (EOQ), we compute the EOQ

\mathbf{EOQ = \sqrt{\frac{2SD}{H}}}

where

D is  demand per year

S is the Ordering cost per order

H is the holding cost per unit

Substituting the values we get,

EOQ = \sqrt{\frac{2*(405*12)*119}{35}}

EOQ = \sqrt{33048}

\mathbf{EOQ = 181.7910889 \approx 182 units}

2. The annual costs of ordering, purchasing and holding the integrated chips is the sum of the cost of ordering, purchasing and holding the integrated  chips.

Since the EOQ at 182 units falls in the second slab of Rich Blue Manufacturing, Bell computer can purchase chips at $325 per unit

Cost of purchasing the chips     \mathbf{405*12*325 = 1579500}

Number of orders to placed         \frac{Annual Demand}{EOQ}

Number of orders to placed         \frac{405*12}{182}

Number of orders to placed   \mathbf{26.703 \approx 27 orders}

Cost of orders                           Number of orders * Cost per order

Cost of orders                               \mathbf{27 * 119 = 3213
}

Holding Costs                        \frac{EOQ}{2} * Holding cost per unit

Holding Costs                               \frac{182}{2} * 35

Holding Costs                               \mathbf{3185
}

Total annual costs       \mathbf{1579500 + 3213 + 3185 = 1585898}

When holding costs are 10% of purchase price per unit,

1. We need to calculate the EOQ, which holding cost at each purchase price

EOQ_{350} = (2*(405*12)*119)/(350*0.10) \approx 182 units

\mathbf{EOQ_{325} = (2*(405*12)*119)/(325*0.10)\approx 189 units}

EOQ_{300} = (2*(405*12)*119)/(300*0.10)\approx 197 units

Since the EOQ lies between 100 and 199 units in all the three costs, Bell Computers can purchase the units only at $325 per unit, so its holding cost will be 10% of $325, which is $32.50 per unit.

2.2. The annual costs of ordering, purchasing and holding the integrated chips is the sum of the cost of ordering, purchasing and holding the integrated  chips.

Since the EOQ at 189 units falls in the second slab of Rich Blue Manufacturing, Bell computer can purchase chips at $325 per unit

Cost of purchasing the chips     \mathbf{405*12*325 = 1579500}

Number of orders to placed         \frac{Annual Demand}{EOQ}

Number of orders to placed         \frac{405*12}{189}

Number of orders to placed   \mathbf{24.762 \approx 25 orders}

Cost of orders                           Number of orders * Cost per order

Cost of orders                               \mathbf{25 * 119 = 2975 }

Holding Costs                        \frac{EOQ}{2} * Holding cost per unit

Holding Costs                               \frac{189}{2} * 32.5

Holding Costs                               \mathbf{3071.25
}

Total annual costs    \mathbf{1579500 + 2975 + 3071.25 = 1585546.25
}

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