# 3) Let K(\theta)=n log (\theta)+c(\theta-1) where c less than 0 and n greater than 0 are known...

## Question:

3) Let {eq}K(\theta)=n log (\theta)+c(\theta-1){/eq} where {eq}c < 0{/eq} and {eq}n > 0{/eq} are known constants. (In

this course, {eq}log (t)= In(t)= log_e(t){/eq}, the natural logarithm of {eq}t{/eq}.) To find the global

maximizer {eq}\theta_M{/eq} of {eq}K(\theta){/eq}, use the following steps.

a) Find {eq}\frac{d}{d \theta} K(\theta){/eq}, set the resulting derivative equal to zero, and solve for {eq}\theta{/eq}.

b) Show that {eq}\frac{d^2}{d \theta ^2} K(\theta) < 0{/eq}.

4) Find {eq}\int _0^3 y^2 \frac{2}{9} ydy{/eq}.

## Critical Points and the Second Derivative Test:

The locations of the critical points of a function {eq}f(x) {/eq} are the solutions to the equation {eq}f'(x) = 0 {/eq}. To classify the identified critical points, we use the second derivative test. In this test, we evaluate the second derivative at each critical point. If the value of the second derivative is positive, the critical point corresponds to a minimum. If the value of the second derivative is negative, the critical point corresponds to a maximum.

3) Let {eq}K(\theta)=n log (\theta)+c(\theta-1){/eq} where {eq}c < 0{/eq} and {eq}n > 0{/eq} are known constants. (In

this course, {eq}log (t)= In(t)= log_e(t){/eq}, the natural logarithm of {eq}t{/eq}.) To find the global

maximizer {eq}\theta_M{/eq} of {eq}K(\theta){/eq}, use the following steps.

{eq}\\ {/eq}

a) Find {eq}\frac{d}{d \theta} K(\theta){/eq}, set the resulting derivative equal to zero, and solve for {eq}\theta{/eq}.

To find the derivative of the function with respect to {eq}\theta {/eq}, we will use the derivative rules {eq}d(\ln(x)) = \frac{1}{x} {/eq} and {eq}d(ax) = x {/eq}.

\begin{align*} K(\theta) &= n\log(\theta) + c(\theta - 1) \\ K'(\theta) &= n(\frac{1}{\theta}) + c \\ K'(\theta) &= \frac{n}{\theta} + c \end{align*}

Now that we have the derivative equation, we can find the critical points by setting the derivative equal to {eq}0 {/eq} and solving for the values of {eq}\theta {/eq} that satisfy the equation.

\begin{align*} K'(\theta) &= \frac{n}{\theta} + c \\ 0 &= \frac{n}{\theta} + c \\ \frac{n}{\theta} &= - c \\ n &= -c\theta \\ \theta &= -\frac{n}{c} \end{align*}

The critical point of the function occurs when {eq}\theta = -\frac{n}{c} {/eq}.

{eq}\\ {/eq}

b) Show that {eq}\frac{d^2}{d \theta ^2} K(\theta) < 0{/eq}.

We must first find the second derivative of the function.

\begin{align*} K(\theta) &= n\ln(\theta) + c(\theta - 1) \\ K'(\theta) &= \frac{n}{\theta} + c \\ K''(\theta) &= -\frac{n}{\theta^2} \end{align*}

We now have an equation for the second derivative of {eq}K {/eq}. Notice, the only appearance of the variable {eq}\theta {/eq} is {eq}\theta^2 {/eq}. This means that for any input, {eq}\theta^2 > 0 {/eq} (since {eq}\theta {/eq} is in the denominator, it cannot equal {eq}0 {/eq}). However, since the numerator is a negative constant, the value of the second derivative will always be negative.

{eq}\\ {/eq}

4) Find {eq}\int _0^3 y^2 \frac{2}{9} ydy {/eq}.

To evaluate the integral, we will first simply. To do this, we'll pull the constant out of the integral and complete the multiplication.

$$\int_0^3 y^2 \frac{2}{9}y \ dy \\ \frac{2}{9} \int_0^3 y^3 \ dy$$

Next, we integrate {eq}y^3 {/eq} with respect to {eq}y {/eq}. Then, we evaluate the identified function at {eq}y = 3 {/eq} and {eq}y = 0 {/eq} and find the difference.

\begin{align*} &= \frac{2}{9} \int_0^3 y^3 \ dy \\ &= \frac{2}{9} (\frac{y^4}{4}) \mid_0^3 \\ &= \frac{2}{9}(\frac{3^4}{4} - \frac{0^4}{4}) \\ &= \frac{2}{9}(\frac{81}{4}) \\ &= \frac{9}{2} \end{align*}

Therefore, {eq}\int_0^3 y^2\frac{2}{9} y \ dy = \frac{9}{2} {/eq}.