Applications of Derivatives
Rate of change
Grade 12
Question:
<p>Given that \(\dfrac{dP}{dx} = 100 - 12\sqrt{x}\). The new level of production of items is \(\displaystyle\int_{2000}^{P} dP = \int_{0}^{25} (100 - 12\sqrt{x})\,dx\). Find the value of \(P\).</p>
<p>2500</p>
<p>3500</p>
<p>4000</p>
<p>3000</p>
Step-by-Step Solution
Key Concept: Recognize that dP/dx represents the rate of change of profit with respect to production, and integrating both sides from their respective limits gives you the change in P. The left side integral directly yields P - 2000, which must equal the right side integral.
<p><strong>Step 1:</strong> Integrate the left side: ∫₂₀₀₀^P dP = P - 2000</p><p><strong>Step 2:</strong> Integrate the right side: ∫₀²⁵ (100 - 12√x)dx = ∫₀²⁵ (100 - 12x^(1/2))dx</p><p>= [100x - 12·(x^(3/2))/(3/2)]₀²⁵ = [100x - 8x^(3/2)]₀²⁵</p><p><strong>Step 3:</strong> Evaluate at the limits:</p><p>= [100(25) - 8(25)^(3/2)] - [0]</p><p>= 2500 - 8(125) = 2500 - 1000 = 1500</p><p><strong>Step 4:</strong> Solve for P: P - 2000 = 1500</p><p>∴ P = 3500</p>
Correct Answer: B