Problem

A manufacturer has determined that the weekly profit from the sale of $x$ items is given by the function below. It is estimated that after $t$ days in any week, $x$ items will have been produced. Find the rate of change of profit with respect to time at the end of 6 days.
\[
P(x)=-x^{2}+600 x-4000 \text { with } x=1.5 t^{2}+4 t
\]

Answer

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Answer

So, the rate of change of profit with respect to time at the end of 6 days is \(\boxed{9768}\)

Steps

Step 1 :Given the profit function \(P(x) = -x^{2} + 600x - 4000\) and the production function \(x = 1.5t^{2} + 4t\)

Step 2 :We need to find the derivative of \(P(x)\) with respect to \(x\), which is \(dP/dx = 600 - 2x\)

Step 3 :Next, we find the derivative of \(x(t)\) with respect to \(t\), which is \(dx/dt = 3.0t + 4\)

Step 4 :Then, we multiply these two derivatives together to find the derivative of \(P\) with respect to \(t\), which is \(dP/dt = (3.0t + 4)(-3.0t^{2} - 8t + 600)\)

Step 5 :Finally, we evaluate this derivative at \(t=6\) to find the rate of change of profit at the end of 6 days, which is 9768

Step 6 :So, the rate of change of profit with respect to time at the end of 6 days is \(\boxed{9768}\)

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