The monthly output of a certain product is $$ Q(x)=1800 x^{5 / 2} $$ where $$ x $$ is the capital investment in milions of dollars. Find $$ 6 Q / d x $$, which can be used to estimate the effect on the output if an additional capital investment of $$ \$ 1 $$ million is made. cQ/ex $$ =\square $$

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To determine the effect on the output $$ Q(x) $$ when an additional capital investment of \$1 million is made, we need to compute the derivative of $$ Q(x) $$ with respect to $$ x $$, which represents the rate of change of output with respect to the capital investment.

Given:
$$
Q(x) = 1800 \, x^{5/2}
$$
where $$ x $$ is the capital investment in millions of dollars.

**Step 1: Compute the derivative $$ \frac{dQ}{dx} $$**

$$
\frac{dQ}{dx} = \frac{d}{dx} \left( 1800 \, x^{5/2} \right) = 1800 \times \frac{5}{2} \, x^{3/2} = 4500 \, x^{3/2}
$$

**Step 2: Interpret the Result**

The derivative $$ \frac{dQ}{dx} = 4500 \, x^{3/2} $$ represents the marginal increase in output for each additional million dollars invested in capital. Therefore, if you invest an additional \$1 million, the approximate increase in output $$ \Delta Q $$ can be estimated as:

$$
\Delta Q \approx \frac{dQ}{dx} \times \Delta x = 4500 \, x^{3/2} \times 1 = 4500 \, x^{3/2}
$$

**Final Answer:**

$$
\frac{dQ}{dx} = 4500 \, x^{3/2}
$$

This means that the derivative $$ \frac{dQ}{dx} $$ is $$ 4500 \, x^{3/2} $$, which estimates the effect on the output for an additional capital investment of \$1 million.

**Answer:**  
After simplifying, the derivative is 4500 · x³⁄². Thus,

dQ/dx = 4500 x^(3/2)
The derivative $$ \frac{dQ}{dx} $$ is $$ 4500 \, x^{3/2} $$.

The monthly output of a certain product is \[ Q(x)=1800 x^{5 / 2} \] where \( x \) is the capital investment in milions of dollars. Find \( 6 Q / d x \), which can be used to estimate the effect on the output if an additional capital investment of \( \$ 1 \) million is made. cQ/ex \( =\square \)

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