Given the requirements and assumptions for a common-emitter amplifier, we will determine the values for resistors R3 and R4. Let's break down the information and proceed step by step:

Since the intended connections or roles of $R_3$ and $R_4$ were misunderstood initially as being directly related to the amplification gain (and thus their direct involvement with the output load resistance), more context is needed to accurately calculate their values. If $R_3$ and $R_4$ form a voltage divider for biasing the base, their values depend heavily on the desired base voltage $V_B$, supply voltage $V_{CC}$, and the quiescent base current $I_{BQ}$, which in turn sets the operating point of the transistor.

For specific guidance on choosing these resistors, kindly clarify their circuit roles or correct the approach if they're intended for a different function than initially interpreted.

The purpose of those resistors is for biasing

Given the clarification that R3 and R4 are indeed meant for biasing in a common-emitter amplifier configuration, let's proceed with calculating their values. The crucial points include setting a stable operating point and ensuring the input impedance requirement is met.

Given quiescent collector current $I_{CQ} = 0.4mA$ and transistor beta ($\beta = 200$),

The base-emitter voltage ($V_{BE_{on}}$) is given as 0.7V. Since we want $V_{CEQ} = 5V$ and assuming the supply voltage $V_{CC} = 10V$ (as per V1's voltage), a stable midpoint biasing would require setting the base voltage ($V_B$) to somewhat above $V_{BE_{on}}$, around 1 to 2 volts, to ensure proper transistor operation without going into saturation or cut-off under small signal variation. Let's target $V_B = 1.5V$ for proper biasing and headroom.

Given $V_B = 1.5V$, the voltage across $R_4$ (from base to ground) will be $1.5V$.

The voltage across $R_3$ will be $V_{CC} - V_B = 10V - 1.5V = 8.5V$.

To ensure the base current $I_{BQ}$ flows through $R_4$ with minimal contribution to input impedance, we target as high a $R_4$ value as possible without compromising the bias stability. The voltage across $R_4$ is $1.5V$, and $I_{BQ} = 2\mu A$,

Since $I_{BQ}$ is negligible for calculating $R_3$ due to the high beta value, the voltage drop across $R_3$ intended is $8.5V$. As $I_{BQ}$ is quite small, the current through $R_3$ is approximately equal to the current through $R_4$, which is $2\mu A$. Therefore,

However, such a high value for $R_3$ coupled with $R_4$ could lead to an excessively high input impedance and less stable biasing due to minimal base current ($I_B$) flow. A better approach would adjust $V_B$ and the resistor values for improved stability.

To improve stability, decrease $R_3$ and $R_4$ values while maintaining their ratio as per the targeted $V_B$. Ensuring $R_{in} \geq 5k\Omega$ allows for some flexibility. The base current need only be a small fraction, allowing for more leeway in resistor values.

A common approach for stable biasing and meeting $R_{in}$ requirements might use a Thevenin equivalent resistance seen by the base $R_{th} \approx 5k\Omega$ to $10k\Omega$, obtained by parallel combination of $R_3$ and $R_4$, and adjust $R_3$ and $R_4$ while checking $V_B$ and $I_{BQ}$ to ensure consistency with $I_{CQ}$ and $V_{CEQ}$.

Given the initial calculations, you should reassess the targeting of $V_B$, $R_3$, and $R_4$ for a practical balance between stability, biasing, and input impedance requirements. The suggested starting values are a stepping stone and might need adjustment based on the actual $V_B$ targeted and considering the impedance seen by the base to ensure it meets $R_{in}$ specifications.

Is there a way to use 16kohm resistors for both R3 and R4.

If the design requires using $16k\Omega$ resistors for both $R_3$ and $R_4$ in the biasing network of a common-emitter amplifier, let's analyze how this will impact the bias and performance of the circuit, especially in terms of setting the quiescent point ($V_{CEQ}$ and $I_{CQ}$) and meeting the input impedance requirement.

The resistors $R_3$ and $R_4$ form a voltage divider that sets the base voltage ($V_B$) to bias the transistor. The voltage divider equation gives:

$V_B = V_{CC} \times \frac{R_4}{R_3 + R_4}$

Given that we're using $16k\Omega$ for both $R_3$ and $R_4$, and assuming $V_{CC} = 10V$ (from the voltage source $V_1$), we have:

$V_B = 10V \times \frac{16k\Omega}{16k\Omega + 16k\Omega} = 10V \times \frac{1}{2} = 5V$

This results in a base voltage of $5V$, which is significantly higher than the typical base-emitter turn-on voltage ($V_{BE_{on}} \approx 0.7V$) for silicon transistors, and it indicates a misunderstanding or an error in the requirement unless there's a specific unconventional design intent that necessitates such a high base voltage, which is unlikely for standard bipolar junction transistor (BJT) operation.

Using $16k\Omega$ resistors for both $R_3$ and $R_4$ without modification or additional components would force an impractical base voltage for a BJT in a common-emitter configuration. We recommend:

If using $16k\Omega$ resistors is a strict requirement due to available parts or other constraints, integrating additional circuit elements to adjust $V_B$ might be necessary for proper transistor operation.