Given the equation below, find when and .
Table of contents
- 0. Functions4h 53m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation2h 18m
- 4. Derivatives of Exponential & Logarithmic Functions1h 16m
- 5. Applications of Derivatives2h 19m
- 6. Graphical Applications of Derivatives6h 0m
- 7. Antiderivatives & Indefinite Integrals48m
- 8. Definite Integrals4h 36m
- 9. Graphical Applications of Integrals1h 43m
- 10. Integrals of Inverse, Exponential, & Logarithmic Functions21m
- 11. Techniques of Integration2h 7m
- 12. Trigonometric Functions6h 54m
- Angles29m
- Trigonometric Functions on Right Triangles1h 8m
- Solving Right Triangles23m
- Trigonometric Functions on the Unit Circle1h 19m
- Graphs of Sine & Cosine46m
- Graphs of Other Trigonometric Functions32m
- Trigonometric Identities52m
- Derivatives of Trig Functions42m
- Integrals of Basic Trig Functions28m
- Integrals of Other Trig Functions10m
- 13: Intro to Differential Equations2h 23m
- 14. Sequences & Series2h 8m
- 15. Power Series2h 19m
- 16. Probability & Calculus45m
5. Applications of Derivatives
Related Rates
Multiple Choice
The perimeter of a rectangle is fixed at 30cm. If the length L is increasing at a rate of 2scm, for what value of L does the area start to decrease? Hint: the rectangle's area starts to decrease when the rate of change for the area is less than 0.
A
L>15cm
B
L>7.5cm
C
L>0cm
D
L>10cm
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Verified step by step guidance1
Step 1: Start by recalling the formula for the perimeter of a rectangle, which is given by P = 2L + 2W, where L is the length and W is the width. Since the perimeter is fixed at 30 cm, we can write 30 = 2L + 2W. Simplify this equation to express W in terms of L: W = 15 - L.
Step 2: Next, recall the formula for the area of a rectangle, which is A = L * W. Substitute W = 15 - L into this formula to express the area in terms of L: A = L * (15 - L). Simplify this to get A = 15L - L^2.
Step 3: To determine when the area starts to decrease, we need to find the rate of change of the area with respect to time, dA/dt. Use the chain rule: dA/dt = (dA/dL) * (dL/dt). First, compute dA/dL by differentiating A = 15L - L^2 with respect to L: dA/dL = 15 - 2L.
Step 4: Substitute dA/dL = 15 - 2L and the given rate of change of the length, dL/dt = 2 cm/s, into the chain rule formula: dA/dt = (15 - 2L) * 2. Simplify this to get dA/dt = 30 - 4L.
Step 5: The area starts to decrease when dA/dt < 0. Solve the inequality 30 - 4L < 0 to find the critical value of L. Rearrange the inequality to isolate L: 4L > 30, which simplifies to L > 7.5 cm. Thus, the area starts to decrease when L exceeds 7.5 cm.
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