Since $x$ appears in the equation after differentiation, we need to find its value by plugging in $p=2$ into the given equation $p^{2}x+6x=5$.

Plug the known values of the variables and their derivatives ($p=2$, $x=1/2$, $\frac{dp}{dt}=3$) into the equation found in Step 3.

Therefore ${\displaystyle \frac{dx}{dt}}=-3/5$ when $p=2$ and ${\displaystyle \frac{dp}{dt}}=3$.

Plug the known values of the variables and their derivatives ($x=300$, $\frac{dx}{dt}=-100$) into the equation found in Step 3.

Therefore revenue is decreasing at $4700 per day when the level of production is 300 cases per day and decreasing at 100 cases per day.

Since $p$ appears in the equation after differentiation, we need to find its value by plugging in $x=4$ into the given demand equation $2p+x+4xp=22$.

Plug the known values of the variables and their derivatives ($x=4$, $p=1$, $\frac{dp}{dt}=25/100$) into the equation found in Step 3.

Therefore

and since $x$ is measured in thousands of units, this means that demand is decreasing by $900$ units per week (i.e., $9/10$ of one thousand) when the weekly demand is 4000 units and the price is increasing at 25 cents per week.

Since $p$ appears in the equation after differentiation, we need to find its value by plugging in $x=20$ into the given supply equation $125p^2-x^2=100$.

Plug the known values of the variables and their derivatives ($x=20$, $p=2$, $\frac{dp}{dt}=-4/100$) into the equation found in Step 3.

Therefore

and since $x$ is measured in thousands of units, this means that supply is falling by $500$ units per week (i.e., $1/2$ of one thousand) when the weekly demand is 20,000 styluses and the price is falling at 4 cents per week.