A price p (in dollars) and demand x (in items) for a product are related by
[tex]2x^2+2xp+50p^2=6200[/tex]If price is increasing at a rate of $2 per month. Therefore,
[tex]\frac{dp}{d\text{ t}}=2[/tex]Now, from the given equation using p=10,
[tex]\begin{gathered} x^2+10x+2500^{}=3100 \\ x^2+10x-600=0 \\ (x+30)(x-20)=0 \\ x=-30,20 \end{gathered}[/tex]Since, demand can't be negative, therefore,
[tex]x=20[/tex]Now, differentiating the given equation w.r.t t and then putting x=20,p=-10,
[tex]\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}[/tex]