ChatGPT vs DeepSeek: Solve a Multi-Step Calculus Optimization Problem

## Solution **Setting up the problem:** Volume constraint: V = πr²h = 500 cm³, so h = 500/(πr²) **Cost function:** - Top + bottom area = 2πr² → cost = 0.06 × 2πr² = 0.12πr² - Side area = 2πrh → cost = 0.03 × 2πrh = 0.06πrh Total cost: C(r) = 0.12πr² + 0.06πrh Substituting h = 500/(πr²): C(r) = 0.12πr² + 0.06πr × 500/(πr²) C(r) = 0.12πr² + 30/r **Finding the minimum:** C'(r) = 0.24πr − 30/r² Setting C'(r) = 0: 0.24πr = 30/r² 0.24πr³ = 30 r³ = 30/(0.24π) = 125/π r = (125/π)^(1/3) ≈ 3.414 cm **Height:** h = 500/(π × 3.414²) = 500/(π × 11.655) = 500/36.617 ≈ 13.655 cm **Second derivative test:** C''(r) = 0.24π + 60/r³ Since both terms are positive for r > 0, C''(r) > 0, confirming this is a minimum. **Verification:** - Volume: π(3.414)²(13.655) = π(11.655)(13.655) = 500.0 ✓ - Cost: 0.12π(11.655) + 30/3.414 = 4.393 + 8.787 = $13.18 - h/r ratio = 13.655/3.414 = 4.0, which makes sense since the side material is cheaper (ratio deviates from h = 2r for equal costs) **Answer:** r ≈ 3.414 cm, h ≈ 13.655 cm, minimum cost ≈ $13.18

## Solution **Step 1: Define variables and constraints** Let r = radius of the can (cm), h = height of the can (cm). Volume: πr²h = 500 → h = 500/(πr²) **Step 2: Build the cost function** The can has two circular ends and a rectangular side when unrolled: - Cost of top + bottom: 2 × πr² × $0.06 = 0.12πr² - Cost of side: 2πrh × $0.03 = 0.06πrh Total cost as a function of r: C(r) = 0.12πr² + 0.06πr · [500/(πr²)] = 0.12πr² + 30r⁻¹ **Step 3: Differentiate and solve** dC/dr = 0.24πr − 30r⁻² Set dC/dr = 0: 0.24πr = 30/r² r³ = 30/(0.24π) = 125/π ∴ r = ∛(125/π) = 5/∛π ≈ 5/1.4646 ≈ 3.4139 cm h = 500/(πr²) = 500/(π · 125²/³/π²/³) = 500π²/³/(π · 125²/³) Simplifying: h = 500/(π · (125/π)²/³) = 4 · 5/∛π = 20/∛π ≈ 13.656 cm Notice: h = 4r. This is expected! When end-cap material costs 2× the side material, the optimal ratio shifts from h = 2r (equal costs) to h = 4r. **Step 4: Confirm minimum via second derivative** d²C/dr² = 0.24π + 60r⁻³ For any r > 0, both terms are strictly positive, so d²C/dr² > 0. ✓ The critical point is indeed a global minimum on (0, ∞). **Step 5: Compute minimum cost** C_min = 0.12π(3.4139)² + 30/(3.4139) = 0.12π(11.655) + 8.787 = 4.393 + 8.787 = **$13.18** **Step 6: Physical sanity check** - Volume: π(3.414)²(13.656) ≈ 500.0 cm³ ✓ - The can is tall and narrow (h ≈ 4r) because end-cap material costs twice as much — so the optimizer shrinks the circles and stretches the side. - If costs were equal, we'd get the classic h = 2r; our 2:1 cost ratio doubles that aspect ratio to 4:1. ✓ **Answer:** r ≈ 3.414 cm, h ≈ 13.656 cm, minimum material cost ≈ $13.18.