Please help thank you

Mathematics

1 answer:

Shkiper50 [21]3 years ago

6 0

Answer:

x = 0.25

Step-by-step explanation:

When logs are added together, they are actually multiplied and then the logs taken of the product.

That sentence is actually correct, but you are going to have to read it a couple of times. You might understand it if I actually just solve the problem.

ln(2x) + ln(2) = 0 Combine the two subjects to make 1 ln.

ln (2)(2x) = 0 Now take the antilog

ln(4x) = 0

antilog ln(4x) = e^0 e^0 = 1

4x = 1 See your last problem.

x = 1/4

Now the question is "What's the answer?" It might be 1/4 but I doubt it. A better choice would be x = 1/4 or x = 0.25

I'd try the last one first.

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<h3>Answer: 15x^(7/3) - 8x^(7/4) + x + 9000</h3>

=========================================================

Explanation:

If you know the cost function C(x), to find the marginal cost, we apply the derivative.

Marginal cost = derivative of cost function

Marginal cost = C ' (x)

Since we're given the marginal cost, we'll apply the antiderivative (aka integral) to figure out what C(x) is. This reverses the process described above.

$\text{Cost} = \text{antiderivative of marginal cost}\\\\\displaystyle C(x) = \int \left(35x^{4/3} - 14x^{3/4} + 1\right)dx\\\\$

$C(x) = \frac{1}{1+4/3}*35x^{4/3+1} - \frac{1}{1+3/4}*14x^{3/4+1} + x + D\\\\C(x) = \frac{1}{7/3}*35x^{7/3} - \frac{1}{7/4}*14x^{7/4} + x + D\\\\C(x) = \frac{3}{7}*35x^{7/3} - \frac{4}{7}*14x^{7/4} + x + D\\\\C(x) = 15x^{7/3} - 8x^{7/4} + x + D\\\\$

D represents a fixed constant. I would have used C as the constant of integration, but it's already taken by the cost function C(x).

To determine the value of D, we plug in x = 0 and C(x) = 9000. This is because we're told the fixed costs are $9000. This means that when x = 0 units are made, you still have $9000 in costs to pay. This is the initial value. You'll find that all of this leads to D = 9000 because everything else zeros out.

Therefore, we go from this

$C(x) = 15x^{7/3} - 8x^{7/4} + x + D\\\\$