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Jlenok [28]
3 years ago
8

A line is represented by the equation 2x + 3y = 15. What is another way to represent the same line, for solving Y ? Help is need

ed please answer with work

Mathematics
1 answer:
Lady bird [3.3K]3 years ago
7 0

The another way to represent the same line, for solving y will be:  y= 5-\frac{2}{3}x

<u><em>Explanation</em></u>

Given equation of the line is:  2x+3y=15

"Solving y" means <u>we need to get</u> y <u>alone in left side.</u> So, we will eliminate all other terms except y from the left side.

First we will <u>subtract 2x </u>from both sides. So, we will get......

2x+3y-2x=15-2x\\ \\ 3y=15-2x

Now, we will <u>divide both sides by 3</u> for getting y alone. So.....

\frac{3y}{3}= \frac{15-2x}{3}\\ \\ y= 5-\frac{2}{3}x

Thus, the another way to represent the same line, for solving y will be:  y= 5-\frac{2}{3}x

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A bakery finds that the price they can sell cakes is given by the function p = 580 − 10x where x is the number of cakes sold per
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Answer:

A) Revenue function = R(x) = (580x - 10x²)

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Step-by-step explanation:

Given,

The Price function for the cake = p = 580 - 10x

where x = number of cakes sold per day.

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

where x = number of cakes sold per day.

Please note that all the calculations and functions obtained are done on a per day basis.

A) Find the revenue and marginal revenue functions [Hint: revenue is price multiplied by quantity i.e. revenue = price × quantity]

Revenue = R(x) = price × quantity = p × x

= (580 - 10x) × x = (580x - 10x²)

Marginal Revenue = (dR/dx)

= (d/dx) (580x - 10x²)

= (580 - 20x)

B) Find the fixed cost and marginal cost function [Hint: fixed cost does not change with quantity produced]

The total cost function is given as

C = (30 + 5x)² = (900 + 300x + 25x²)

The total cost function is a sum of the fixed cost and the variable cost.

The fixed cost is the unchanging part of the total cost function with changing levels of production (quantity produced), which is the term independent of x.

C(x) = 900 + 300x + 25x²

The only term independent of x is 900.

Hence, the fixed cost = 900

Marginal Cost function = (dC/dx)

= (d/dx) (900 + 300x + 25x²)

= (300 + 50x)

C) Find the profit function [Hint: profit is revenue minus total cost]

Profit = Revenue - Total Cost

Revenue = (580x - 10x²)

Total Cost = (900 + 300x + 25x²)

Profit = P(x)

= (580x - 10x²) - (900 + 300x + 25x²)

= 580x - 10x² - 900 - 300x - 25x²

= 280x - 35x² - 900

= (-35x² + 280x - 900)

D) Find the quantity that maximizes profit

To obtain this, we use differentiation analysis to obtain the maximum point of the Profit function.

At maximum point, (dP/dx) = 0 and (d²P/dx²) < 0

P(x) = (-35x² + 280x - 900)

(dP/dx) = -70x + 280 = 0

70x = 280

x = (280/70) = 4

(d²P/dx²) = -70 < 0

Hence, the point obtained truly corresponds to a maximum point of the profit function, P(x).

This quantity demanded obtained, is the quantity demanded that maximises the Profit function.

Hope this Helps!!!

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