A price p (in dollars) and demand x (in items) for a product are related by
![2x^2+2xp+50p^2=6200](https://img.qammunity.org/2023/formulas/mathematics/college/uuab4bhk5x64r6t04lzj5xlo4p0tb98fi5.png)
If price is increasing at a rate of $2 per month. Therefore,
![\frac{dp}{d\text{ t}}=2](https://img.qammunity.org/2023/formulas/mathematics/college/zso5lifvrxw7nev2liszgnir82hpmq3q4w.png)
Now, from the given equation using p=10,
![\begin{gathered} x^2+10x+2500^{}=3100 \\ x^2+10x-600=0 \\ (x+30)(x-20)=0 \\ x=-30,20 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/icua5oppztahmmufox71vzw9k3fy4caiyo.png)
Since, demand can't be negative, therefore,
![x=20](https://img.qammunity.org/2023/formulas/mathematics/high-school/otvww2o09b1qyegcnswgop1y82pbf6uftp.png)
Now, differentiating the given equation w.r.t t and then putting x=20,p=-10,
![\begin{gathered} 4xx^(\prime)+2xp^(\prime)+2px^(\prime)+100pp^(\prime)=0 \\ 4(20)x^(\prime)+2(20)(2)+2(10)x^(\prime)+100(10)(2)=0 \\ 80x^(\prime)+80+20x^(\prime)+1000=0 \\ 100x^(\prime)=-1040 \\ x^(\prime)=-10.40 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fcwqg29uaibiud4qhjvfoh44100r4krja6.png)