49:I[79858,["1216","static/chunks/eeac573e-cfb1e84259e7dde2.js","4649","static/chunks/4649-d659720c90d9ee35.js","6404","static/chunks/6404-6ef05b8bff2887b0.js","4797","static/chunks/4797-e81dc81e0c1c1336.js","2972","static/chunks/2972-ab992f8c79eda392.js","4296","static/chunks/4296-8f39ea5654a4f1ab.js","4235","static/chunks/4235-6fd3b88ee1c357a0.js","960","static/chunks/960-4945c17e7bdffd04.js","866","static/chunks/866-36e9a2593ae60457.js","6501","static/chunks/app/%5BsubjectLabel%5D/problems/%5BproblemHash%5D/page-756d64521be57e67.js"],"ProblemPageMemo"] 4a:T761,

Аналитическое:

Пусть P_d = a - bQ, AC = c. Тогда TC = cQ. По условию, b=1.

MR = a - 2Q = MC = c \Rightarrow Q^* = \frac{a - c}{2}

\pi_{\text{max}} = \left( a - \frac{a - c}{2} \right) \cdot \frac{a - c}{2} - c \cdot \frac{a - c}{2} = \left(\frac{a - c}{2}\right)^2 = \left( Q^* \right)^2

Q^* = \sqrt{\pi_{\text{max}}} = 2009

Геометрическое:

Наклон кривой спроса равен 1. Поэтому

\begin{cases} OF = OE \\ OB = BE \\ \frac{OF}{OB} = 2. \end{cases}

В силу подобия \triangle FOB и \triangle FCA,

\frac{FC}{CA} = \frac{FO}{OB} = 2.

Получаем:

\pi_{\text{max}} = S_{\triangle FCA} = 0,5 \cdot CA \cdot FC = 0,5 \cdot