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3 years ago

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A camera manufacturer spends $2,100 each day for overhead expenses plus $9 per camera for labor and materials. The cameras sell

for $14 each. a. How many cameras must the company sell in one day to equal its daily costs? b. If the manufacturer can increase production by 50 cameras per day, what would their daily profit be?

2 answers:

alexandr402 [8]3 years ago

4 0

Here, first we need to find the number of cameras that can be sold, so that total costs are equal to total revenues.

Let the number of cameras that can be sold be "x".

Total revenue = $ 14x

Total costs will be = $ 2100 + $9x

So, the equation that will be formed ⇒

14 x = 2100 + 9x

5x = 2100

x = 420

<u>Thus, in order to make total costs equal to total revenues, 420 cameras are required to be sold.</u>

If instead of 420, 470 cameras are sold, the daily profit will be calculated as under -

Total revenue = $ 14 × 470 cameras = $ 6,580

Total costs = $ 2100 + ( $ 9 × 470 cameras) = $ 6,330

<u>Daily profit = Total revenue - Total costs = $ 6,580 - $ 6,330 = $ 250</u>

alexdok [17]3 years ago

3 0

$<span>400 have a nice one! :) </span>

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4 years ago

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Saddle point: $(0,0)$

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Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

Step-by-step explanation:

The function is:

$f(x,y) = 8\cdot y^{2}\cdot x -8\cdot y\cdot x^{2} + 9\cdot x \cdot y$

The partial derivatives of the function are included below:

$\frac{\partial f}{\partial x} = 8\cdot y^{2}-16\cdot y\cdot x+9\cdot y$

$\frac{\partial f}{\partial x} = y \cdot (8\cdot y -16\cdot x + 9)$

$\frac{\partial f}{\partial y} = 16\cdot y \cdot x - 8 \cdot x^{2} + 9\cdot x$

$\frac{\partial f}{\partial y} = x \cdot (16\cdot y - 8\cdot x + 9)$

Local minima, local maxima and saddle points are determined by equalizing both partial derivatives to zero.

$y \cdot (8\cdot y -16\cdot x + 9) = 0$

$x \cdot (16\cdot y - 8\cdot x + 9) = 0$

It is quite evident that one point is (0,0). Another point is found by solving the following system of linear equations:

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The solution of the system is (3/8, -3/8).

Let assume that y = 0, the nonlinear system is reduced to a sole expression:

$x\cdot (-8\cdot x + 9) = 0$

Another solution is (9/8,0).

Now, let consider that x = 0, the nonlinear system is now reduced to this:

$y\cdot (8\cdot y+9) = 0$

Another solution is (0, -9/8).

The next step is to determine whether point is a local maximum, a local minimum or a saddle point. The second derivative test:

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The second derivatives of the function are:

$\frac{\partial^{2} f}{\partial x^{2}} = 0$

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$\frac{\partial^{2} f}{\partial x \partial y} = 16\cdot y -16\cdot x + 9$

Then, the expression is simplified to this and each point is tested:

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S2: (3/8,-3/8)

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S3: (9/8, 0)

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S4: (0, - 9/8)

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Unfortunately, the second derivative test associated with the function does offer an effective method to distinguish between local maximum and local minimums. A more direct approach is used to make a fair classification:

S2: (3/8,-3/8)

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S4: (0, - 9/8)

$f(0,-\frac{9}{8} ) = 0$ (Local maximum)

Saddle point: $(0,0)$

Local minimum: $(\frac{3}{8}, -\frac{3}{8})$

Local maxima: $(0,-\frac{9}{8})$ , $(\frac{9}{8},0)$

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