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matrenka [14]
3 years ago
10

Grant thinks he has reached the point where he has maximized his profit. However, because this can be tricky to determine, he is

not sure. He decides to test it by selling one more unit. If Grant is correct in his assumption,
what should happen when he sells this additional unit?
a. Grant will break even on his sales
b. Profit will increase but at a slower rate
c. Marginal revenue will exceed marginal cost
d. Marginal cost will exceed marginal revenue
e. Profits will be reduced to zero
Mathematics
1 answer:
Mashcka [7]3 years ago
6 0

Answer:c

Step-by-step explanation:

Profit maximization happens with marginal revenue is equal to marginal cost, so if Grant's assumption was right before selling the extra unit, when he actually sells the extra unit, this will increase his revenue

That's why the answer is c

Marginal revenue will exceed marginal cost

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Evaluate the following double integral where a = 2y
Keith_Richards [23]

Change the order of integration.

\displaystyle \int_0^1 \int_{2y}^2 \cos(x^2) \, dx \, dy = \int_0^2 \int_0^{x/2} \cos(x^2) \, dy \, dx \\\\ ~~~~~~~~ = \int_0^2 \cos(x^2) y \bigg|_{y=0}^{y=x/2} \, dx \\\\ ~~~~~~~~ = \frac12 \int_0^2 x \cos(x^2) \, dx

Substitute u=x^2 and du=2x\,dx.

\displaystyle \frac12 \int_0^2 x \cos(x^2) \, dx = \frac14 \int_0^4 \cos(u) \, du = \frac14 \sin(u) \bigg|_{u=0}^{u=4} = \boxed{\frac{\sin(4)}4}

4 0
1 year ago
Find the integral using substitution or a formula.
Nadusha1986 [10]
\rm \int \dfrac{x^2+7}{x^2+2x+5}~dx

Derivative of the denominator:
\rm (x^2+2x+5)'=2x+2

Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

Based on our previous test, we know that a simple substitution will work for the first integral: \rm \quad u=x^2+2x+5\qquad\to\qquad du=2x+2~dx

So the first integral changes,

\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

integrating to a log,

\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

\rm x^2+2x+5=(x^2+2x+1)+4=(x+1)^2+2^2

So we have this going on,

\rm ln|x^2+2x+5|+\int\dfrac{5}{(x+1)^2+2^2}~dx

Let's factor the 5 out of the intergral,
and the 4 from the denominator,

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Bringing all that stuff together as a single square,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(\dfrac{x+1}{2}\right)^2+1}~dx

Making the substitution: \rm \quad u=\dfrac{x+1}{2}\qquad\to\qquad 2du=dx

giving us,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\left(u\right)^2+1}~2du

simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

\rm ln|x^2+2x+5|+\frac52arctan(u)

undo your substitution as a final step,
and include a constant of integration,

\rm ln|x^2+2x+5|+\frac52arctan\left(\frac{x+1}{2}\right)+c

Hope that helps!
Lemme know if any steps were too confusing.

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