Mathematics, 06.07.2019 08:30 Pizzapegasus1

Arestaurant sells 330 sandwiches each day. for each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. how much should the restaurant charge to maximize daily revenue? what is the maximum daily revenue? each sandwich is 6$

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The value that optimizes this inequality is x = 1, since it represents the vertex of the parabola. Therefore, $ 0.25 is the price that must be discounted to maximize revenues

With x = 1 the value of the new income will be = $ 1983.75

Step-by-step explanation:

Revenue with the current sale price is:

6 * 330 = $ 1980

Let's call x the number of times the price of a sandwich decreases, that is, if x = 1 then the price of the sandwich decreases by $ 0.25

Then, the price would be:

Price = (6-0.25x)

Then, if the price is decreased by a factor of x, then the number of sandwiches sold will increase by a factor of 15x. Therefore the number of clients will be:

Sale = (330 + 15x)

Now we need an equation for income. The income will be equal to the sale price, for the number of sandwiches sold

Income = price * sale

Income = (6-0.25x) (330 + 15x)

In order for this restaurant pricing model to be profitable, the revenues with the new sale price must be greater than the revenues with the sale price of $ 6. So:

New income> current income.

$(6-0.25x) * (330 + 15x) 330 * 6\\ (6-0.25x) * (330 + 15x) 1980$

The left side of the inequality represents a parabola:

-3,750x ^ 2 + 7.5x + 1980 1980

The value that optimizes this inequality is x = 1, since it represents the vertex of the parabola. Therefore, $ 0.25 is the price that must be discounted to maximize revenues

The attached image shows the region that satisfies the proposed inequality and the vertex of the parabola:

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

With x = 1 the value of the new income will be = $ 1983.75

Arestaurant sells 330 sandwiches each day. for each $0.25 decrease in price, the restaurant sells ab

Answer from: Quest

check the picture below.

well, the angle at the vertex d is clearly not a right-angle, let's check for the one at the vertex f which does look like, however if it's indeed, that means ef is perpendicular to df, in which case their slopes will be negative reciprocal of each other.

$\bf e(\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{0-2}{)}\implies \implies \cfrac{-2}{0+2}\implies \cfrac{-2}{2}\implies -1 \\\\[-0.35em] ~\dotfill$

$\bf d(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-1})\qquad f(\stackrel{x_2}{0}~,~\stackrel{y_2}{0}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{)}{)}\implies \cfrac{0+1}{0+2}\implies \cfrac{1}{2} \\\\[-0.35em] ~\dotfill$

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope~of~ef}{-1\implies \cfrac{-1}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{1}{1}\implies 1}}$

so the negative reciprocal of the slope of ef is 1, however the slope of df is not 1, is 1/2, so nope, they're not perpendicular lines.

Answer from: Quest

The answer is all about the number 4

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Arestaurant sells 330 sandwiches each day. for each $0.25 decrease in price, the restaurant sells ab...