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I think its the second but i'm not sure because there are no measurements

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First derivative of the total profit function is dp/dx = d/dx(-0.001x^2 + 4x - 1850) = -0.002x + 4 when dp/dx is taken equal to 0 - 0.002x + 4 = x = 4/0.002 = 2000 second derivative is d^2p/dx^2 = d/dx(dp/dx) = d/dx(-0.002x + 4) = - 0.002 since the second derivative is - ve, it can be said the profit function has a maximum at x = 2000 profit will be maximum when 2000 cans are sold 1) maximum profit = p(x=2000) = -0.001(2000)^2 + 4*2000 - 1850 = -4000 + 8000 - 1850 = 2150 2) he must sell 2000 cans for getting maximum profit.