Math Problem Statement
where is thousands of units and is thousands of dollars.
(a) What is revenue when 4 thousand units are sold?
Ans: thousand dollars
(b) What is the smallest number of units we need to sell in order to have 361.152 thousand dollars of revenue?
Ans: thousand units
(c) For what value of is revenue maximized. Round your answer to the nearest tenth (one decimal place).
Ans: thousand units
Solution
To solve these questions, we analyze the given revenue function:
where is in thousands of units, and is in thousands of dollars.
(a) What is revenue when 4 thousand units are sold?
Substitute into the revenue function:
Answer (a): thousand dollars.
(b) Smallest for thousand dollars
We solve . Rearrange:
This is a cubic equation. Solving it yields:
Answer (b): thousand units.
(c) Value of for maximum revenue
Revenue is maximized at the critical points, where the derivative of with respect to is zero: Solve for :
Answer (c): thousand units.
Summary:
(a) thousand dollars
(b) thousand units
(c) thousand units
Would you like detailed step-by-step solutions or have any questions?
Here are 5 related questions for practice:
- How does increasing the unit price affect revenue for this function?
- Solve for when revenue is thousand dollars.
- What is the second derivative of the revenue function, and what does it indicate?
- If , what is the new revenue when ?
- How do we calculate the point of inflection for the given function?
Tip: Always check for concavity changes when analyzing maximum or minimum points!
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Math Problem Analysis
Mathematical Concepts
Algebra
Cubic Equations
Calculus
Optimization
Formulas
Revenue function: R(q) = -q^3 + 169q
Critical points: Solve R'(q) = 0
Solving cubic equations: ax^3 + bx^2 + cx + d = 0
Theorems
Derivative for finding critical points
First derivative test for optimization
Suitable Grade Level
Grades 11-12
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