Given the values:
"TC=400+20Q\u22122Q^2+Q^3\\\\AR = MR = \\$180"
a:
"TC=400+20Q\u22122Q^2+Q^3\\\\MC =\\frac{ dTC}{dQ}=20\u22124Q+3Q^2\\\\Now,\\space MC = MR\\\\20\u22124Q+3Q^2=180\\\\0=180 \u2212 (20\u22124Q+3Q^2)\\\\0=180\u221220+4Q\u22123Q^2\\\\0=160+4Q\u22123Q^2"
Now solve for the Q, thus, Q=8 and "Q=\\frac{\u221220}{\n\n3}"
So consider the positive value Q=8.
Thus, profit maximizing level of output is = 8
"Profit = TR \u2212 TC\\\\Profit = (MR\u00d7Q)\u2212(400+20Q\u22122Q^2+Q^3)\\\\Profit = (180\u00d78)\u2212(400+20\u00d78\u22122\u00d78^2+8^3)\\\\Profit = 1440\u2212(400+160\u2212128+512)\\\\Profit = 1440\u2212(944)\\\\Profit = \\$496"
b:
Shutdown point,
"AVC = MC\\\\TC = FC+VC\\\\Thus, \\space VC = 20Q\u22122Q^2+Q^3\\\\AVC=\\frac{VC}{Q}\\\\AVC=\\frac{20Q\u22122Q^2+Q^3}{Q}\\\\AVC=20\u22122Q+Q^2\\\\At\\space shutdown\\space point,\\\\ MC=AVC\\\\20\u22124Q+3Q^2=20\u22122Q+Q^2\\\\20\u22124Q+3Q^2\u2212(20\u22122Q+Q^2)=0\\\\20\u22124Q+3Q^2\u221220+2Q\u2212Q^2=0\\\\\u22122Q+2Q^2=0\\\\2Q(\u22121+Q)=0\\\\\u22121+Q=0\\\\Q=1"
shutdown level of quantity = 1
"P=MC\\\\P=20\u22124Q+3Q^2\\\\P=20\u22124(1)+3(1)^2\\\\P=20\u22124+3\\\\P=\\$19"
shutdown price =$19
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