Given the region bounded by the x-axis, the y-axis, and the line in the first quadrant, determine the and coordinates of the centroid of the shaded area.
Table of contents
- 0. Functions7h 55m
- Introduction to Functions18m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms36m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 31m
- 12. Techniques of Integration7h 41m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
5. Graphical Applications of Derivatives
Applied Optimization
Multiple Choice
Your café sells lattes for \$4 each to 100 customers per day. For every \$1 increase in price, you would lose 20 customers. Find the price that maximizes revenue. Hint: The # of items sold is based on the number of customers.
A
\$4.00
B
\$4.50
C
\$5.00
D
\$5.50
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Verified step by step guidance1
Define the variables: Let x be the number of \$1 increases in price. The new price per latte will be (4 + x) dollars.
Express the number of customers as a function of x: Since you lose 20 customers for each \$1 increase, the number of customers will be (100 - 20x).
Write the revenue function R(x): Revenue is the product of price and quantity sold, so R(x) = (4 + x)(100 - 20x).
Expand the revenue function: Distribute to get R(x) = 400 + 100x - 80x - 20x^2, which simplifies to R(x) = 400 + 20x - 20x^2.
Find the maximum revenue: Since R(x) is a quadratic function opening downwards (coefficient of x^2 is negative), find the vertex using x = -b/(2a) where a = -20 and b = 20. This will give the value of x that maximizes revenue.
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