Question:** A venture capitalist is evaluating a cleantech startup’s cost model given by \( C(x) = 5000 + 120x - 0.5x^2 \), where \( C \) is the cost in dollars for producing \( x \) units. Find the production level \( x \) that minimizes cost per unit.

["Title: Minimizing Cost Per Unit in Cleantech Production: How Venture Capitalists Evaluate Startup Cost Models", "Meta Description:\nWhen evaluating a cleantech startup’s cost model, determining the optimal production level that minimizes cost per unit is critical. Using calculus, we analyze the cost function ( C(x) = 5000 + 120x - 0.5x^2 ) to find the production volume ( x ) that achieves this efficiency.", "---", "### Introduction: The Crucial Link Between Cost and Scaling in Cleantech", "For cleantech startups striving to scale sustainably, understanding unit cost dynamics is essential. High production costs can hinder competitiveness and investment returns—making precise cost modeling a cornerstone of due diligence.", "One common cost function is quadratic, such as ( C(x) = 5000 + 120x - 0.5x^2 ), where ( C(x) ) represents total cost in dollars to produce ( x ) units- Given this model, venture capitalists seek the production level ( x ) that minimizes the cost per unit, a key efficiency metric.", "In this article, we walk through the mathematical optimization to identify the optimal production scale.", "---", "### Understanding the Cost Function: Total Cost and Cost Per Unit", "The given cost function is:\n[\nC(x) = 5000 + 120x - 0.5x^2\n]\nHere:\n- ( 5000 ) is a fixed startup or setup cost\n- ( 120x ) reflects marginal variable production costs\n- ( -0.5x^2 ) introduces decreasing marginal returns, a realistic feature in manufacturing (e.g., resource constraints or inefficiencies at high scale)", "The cost per unit is calculated as:\n[\n\ ext{Cost per unit} = \frac{C(x)}{x} = \frac{5000 + 120x - 0.5x^2}{x}\n]\nSimplify:\n[\n\frac{C(x)}{x} = \frac{5000}{x} + 120 - 0.5x\n]\nWe now define this as a function:\n[\nf(x) = \frac{5000}{x} + 120 - 0.5x\n]\nOur goal is to minimize ( f(x) ) for ( x > 0 ), since production must be positive.", "---", "### Minimizing Cost Per Unit Using Calculus", "To minimize ( f(x) ), we take its derivative and find critical points:\n[\nf'(x) = \frac{d}{dx}\left( \frac{5000}{x} + 120 - 0.5x \right) = -\frac{5000}{x^2} - 0.5\n]", "Wait—this derivative is ( -\frac{5000}{x^2} - 0.5 ), which is always negative for ( x > 0 ). That suggests ( f(x) ) is strictly decreasing, which contradicts the idea of a minimum.", "But this result seems counterintuitive—how can cost per unit decrease forever? It can’t. We made a modeling assumption that requires revision.", "Hold on: Minimizing cost per unit with a quadratic cost function requires careful handling.", "Actually, the distance from ( C(x) ) to ( x \cdot \ ext{avg. cost} ) isn’t directly minimized by simple calculus in this form—because ( f(x) = \frac{C(x)}{x} ) has no minimum when ( C(x) ) is concave down. That signals that we should reevaluate the correct functional form for optimization in operations research.", "---", "### Correcting the Approach: Minimizing Marginal vs. Average Cost", "In economics and operations, minimizers of average cost are found by setting the first derivative of average cost to zero, provided it’s the global minimum.", "While our earlier simplification led to a non-minimizing function, let’s re-evaluate the full cost model from a more conventional perspective.", "Suppose instead the startup’s total cost reflects realistic efficiency:", "[\nC(x) = a + bx - cx^2 \quad (a,b,c > 0)\n]\nThen average cost is:\n[\nAC(x) = \frac{a}{x} + b - cx\n]", "Take derivative:\n[\nAC'(x) = -\frac{a}{x^2} - c\n]\nSet ( AC'(x) = 0 ):\n[\n-\frac{a}{x^2} - c = 0 \quad \Rightarrow \quad \frac{a}{x^2} = -c\n]\nBut this implies a negative derivative—so average cost decreases forever if ( c < 0 )? No—this suggests a flaw in sign conventions.", "Wait: ( c > 0 ) implies declining marginal cost initially, but infinite growth later. Average cost ( AC(x) ) typically has a minimum when:\n[\nAC'(x) = -\frac{a}{x^2} - c < 0 \quad \ ext{for all } x > 0 \Rightarrow \ ext{AC}(x) \ ext{ is strictly decreasing}\n]\nBut that contradicts intuition—no real-world cleantech plant should have ever-decreasing average cost.", "Ah—this suggests the model may be inverted. In standard economic models, cost per unit decreases at first, then increases. The correct second-order model includes a diminishing marginal return, so:\n[\nC(x) = C_0 + \alpha x - \beta x^2, \quad \beta > 0\n]\nBut again, derivative\n[\nC'(x) = \alpha - 2\beta x\n\Rightarrow \ ext{Minimum at } x = \frac{\alpha}{2\beta}\n]", "So where does our original ( C(x) = 5000 + 120x - 0.5x^2 ) fit? This function has negative derivative for all ( x > 0 ), meaning cost per unit always decreases—impossible in real production.", "This implies the model likely lacks fixed costs contributing to average cost plateaus—or perhaps the optimization objective is misstated.", "---", "### Realizing the Optimal Production Level in Such Models", "Given ( C(x) = 5000 + 120x - 0.5x^2 ), compute cost per unit:\n[\nf(x) = \frac{5000}{x} + 120 - 0.5x\n]\nEven though derivative ( f'(x) = -\frac{5000}{x^2} - 0.5 < 0 ), which decreases always, real-world analysis shows that average cost has a unique minimum when total cost is neither always increasing nor decreasing—but this model lacks convexity.", "However, suppose the startup’s true model is:\n[\nC(x) = a x + b x^2 \quad (b < 0)\quad \ ext{i.e., quadratic margin cost}\n]\nThen with ( b < 0 ), average cost has a single minimum. But our case has ( -0.5x^2 ), so ( b = -0.5 < 0 ), but coefficient of ( x ) is positive—still, the model’s concavity matters.", "Let’s plot } ( f(x) = \frac{5000}{x} + 120 - 0.5x )\n- At ( x = 10 ): ( f(10) = 500 + 120 - 5 = 615 )\n- At ( x = 20 ): ( f(20) = 250 + 120 - 10 = 360 )\n- At ( x = 30 ): ( f(30) = 166.7 + 120 - 15 = 271.7 )\n- At ( x = 40 ): ( f(40) = 125 + 120 - 20 = 225 )\n- At ( x = 50 ): ( f(50) = 100 + 120 - 25 = 195 )\n- At ( x = 60 ): ( f(60) = 83.3 + 120 - 30 = 173.3 )\n- At ( x = 70 ): ( f(70) = 71.4 + 120 - 35 = 156.4 )\n- At ( x = 80 ): ( f(80) = 62.5 + 120 - 40 = 142.5 )\n- At ( x = 90 ): ( f(90) = 55.6 + 120 - 45 = 130.6 )\n- At ( x = 100 ): ( f(100) = 50 + 120 - 50 = 120 )\n- At ( x = 110 ): ( f(110) = 45.45 + 120 - 55 = 110.45 )\n- At ( x = 120 ): ( f(120) = 41.67 + 120 - 60 = 101.67 )\n- At ( x = 130 ): ( f(130) = 38.46 + 120 - 65 = 93.46 )\n- At ( x = 140 ): ( f(140) = 35.71 + 120 - 70 = 85.71 )\n- At ( x = 150 ): ( f(150) = 33.33 + 120 - 75 = 78.33 )\n- At ( x = 160 ): ( f(160) = 31.25 + 120 - 80 = 71.25 )\n- At ( x = 170 ): ( f(170) = 29.41 + 120 - 85 = 54.41 )\n- At ( x = 180 ): ( f(180) = 27.78 + 120 - 90 = 57.78 )", "Cost per unit dips around ( x = 170 ), but at ( x = 160 ), ( f(x) = 71.25 ); ( x = 170 ), 54.41; ( x = 180 ), 57.78 — minimum near 170? But derivative never zero.", "However, the minimum of ( f(x) ) occurs at ( x \ o \infty )? No—derivative remains negative.", "Wait: derivative is ( f'(x) = -\frac{5000}{x^2} - 0.5 < 0 ), so function is strictly decreasing for all ( x > 0 "]