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anastassius [24]

3 years ago

5

A small business makes 3-speed and 10-speed bicycles at two different factories. Factory A produces 16 3-speed and 20 10-speed b

ikes in one day while factory B produces 12 3-speed and 20 10- speed bikes daily. It costs $1000/day to operate factory A and $800/day to operate factory B. An order for 80 3-speed bikes and 140 10-speed bikes has just arrived How many days should each factory be operated in order to fill this order at a minimum cost? (Give your answers correct to two decimal places.) Factory A should be operated Factory B should be operated days. days. What is the minimum cost? (Give your answer correct to the nearest dollar.)

1 answer:

GarryVolchara [31]3 years ago

6 0

Answer:

Factory A should be operated 6.11 days

Factory B should be operated 6.88 days

The minimum cost for factory A is $6,110

The minimum cost for factory B is $5,504

Step-by-step explanation:

Factory A

Daily production: 16 3-speed and 20 10-speed bikes

Total daily production = 16+20 = 36 speed bikes

Order: 80 3-speed and 140 10-speed bikes

Total order = 80+140 = 220 speed bikes

Number of days = 220/36 = 6.11 days

Cost per day = $1,000

Minimum cost for 6.11 days = $1000 × 6.11 = $6,110

Factory B

Daily production: 12 3-speed and 20 10-speed bikes

Total daily production = 12+20 = 32 speed bikes

Total order = 220 speed bikes

Number of days = 220/36 = 6.88 days

Cost per day = $800

Cost for 6.88 days = $800 × 6.88 = $5,504

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Consider the equation below. f(x) = 2x3 + 3x2 − 336x (a) Find the interval on which f is increasing. (Enter your answer in inter

Vaselesa [24]

We have been given a function $f(x)=2x^3+3x^2-336x$ . We are asked to find the interval on which function is increasing and decreasing.

(a). First of all, we will find the critical points of function by equating derivative with 0.

$f'(x)=2(3)x^{2}+3(2)x^1-336$

$f'(x)=6x^{2}+6x-336$

$6x^{2}+6x-336=0$

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$x=-8,x=7$

So $x=-8,7$ are critical points and these will divide our function in 3 intervals $(-\infty,-8)U(-8,7)U(7,\infty)$ .

Now we will find derivative over each interval as:

$f'(x)=(x+8)(x-7)$

$f'(-9)=(-9+8)(-9-7)=(-1)(-16)=16$

Since $f'(9)>0$ , therefore, function is increasing on interval $(-\infty,-8)$ .

$f'(x)=(x+8)(x-7)$

$f'(1)=(1+8)(1-7)=(9)(-6)=-54$

Since $f'(1)$ , therefore, function is decreasing on interval $(-8,7)$ .

Let us check for the derivative at $x=8$ .

$f'(x)=(x+8)(x-7)$

$f'(8)=(8+8)(8-7)=(16)(1)=16$

Since $f'(8)>0$ , therefore, function is increasing on interval $(7,\infty)$ .

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Let us find values of f(x) on these two points.

$f(-8)=2(-8)^3+3(-8)^2-336(-8)$

$f(-8)=1856$

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$f(7)=-1519$

Therefore, $(-8,1856)$ is a local maximum and $(7,-1519)$ is a local minimum.

(c) To find inflection points, we need to check where 2nd derivative is equal to 0.

Let us find 2nd derivative.

$f''(x)=6(2)x^{1}+6$

$f''(x)=12x+6$

$12x+6=0$

$12x=-6$

$\frac{12x}{12}=-\frac{6}{12}$

$x=-\frac{1}{2}$

Therefore, $x=-\frac{1}{2}$ is an inflection point of given function.

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3 years ago

2) Evaluate 9 X 364 – 9 X 125 – 9 X 39 without the use of your calculator

Brums [2.3K]

Answer:

<em>9*364 - 9*125 - 9*39 = 1,800</em>

Step-by-step explanation:

<em>Mental Calculation</em>

If we detect known patterns in the calculations, we could easily give their results without the use of calculators.

We are given the expression to evaluate:

9*364 - 9*125 - 9*39

The first thing to note is the 9 is a common factor of all terms, thus we take it out:

9*(364 - 125 - 39)

The negative numbers can be easily added

125

+ 39

------------

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Now our expression is much easier:

9*(364 - 164)

The subtraction of 364-164 is 200, thus the result of the operations is:

9*200. We only need to recall 9*2=18 and add two zeros to get 1,800, thus:

9*364 - 9*125 - 9*39 = 1,800

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