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3 years ago

If fixed costs are 100, and variable costs are at 5 per unit, find the total cost when Q=10.

Mathematics

1 answer:

kompoz [17]3 years ago

6 0

The total cost when Q = 10 is $150

Step-by-step explanation:

Th equation y = mx + b represents any linear function, where

m is the unit rate
b is the fixed amount

∵ The fixed costs are $100

∵ The variable costs are $5 per unit

∵ There are Q units

- Represent the total cost by the function c(Q)

∴ c(Q) = 5 Q + 100, where c(Q) is the total cost of Q units

∵ c(Q) = 5 Q + 100

∵ Q = 10

- Substitute Q by 10

∴ c(10) = 5(10) + 100

∴ c(10) = 50 + 100

∴ c(10) = 150

The total cost when Q = 10 is $150

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A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

$=30x^2+6xh$

Let

C $=30x^2+6xh$

Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

The length of one side of the base of the box is 3 ft.

The height of the box is $=\frac{272}{3^2}$

=30.22 ft.

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

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3 years ago

which rate can you set 7 miles over 1 hour equal to in order to find the distance traveled in 49 hours at 7 miles per hour

IrinaVladis [17]

Answer:

Step-by-step explanation:

time = 49 hours

speed = 7 miles/hour

speed = distance / time

∴ distance = speed × time

= 7 × 49

= 343 miles

8 0

3 years ago

If 2x − 4(x + 1) = −12, evaluate x2 − 1.

Vlad [161]

2x-4(x+1)=-13
1) we solve this equation:
2x-4(x+1)=-12
2x-4x-4=-12
-2x=-12+4
-2x=-8
x=-8/-2
x=4

2) we evalute x²-1
x²-1=4²-1=16-1=15

Answer: 15

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3 years ago

If x = 5, find the value of<br> a) x2<br> b) (x + 3)2

PtichkaEL [24]

Answer: a) = 10

b) = 16

Step-by-step explanation:

a) as x = 5 so putting the value of x in given equation

5 multiplied by 2 is equal to 10

b) Putting value of x in given equation

= (5 + 3) 2

According to BODMAS solve the brackets first

= 8 x 2

= 16

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Which is a zero if the quadratic function f(x) = 16x2 + 32x - 9

Artist 52 [7]

x= 1/4 , -9/4

when you set the function to equal to zero you will get these values

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