Math Problem Statement

<math xmlns="http://www.w3.org/1998/Math/MathML"> <mstyle displaystyle="true"> <mi>R</mi> <mrow> <mo>(</mo> <mi>q</mi> <mo>)</mo> </mrow> <mo>=</mo> <mo>-</mo> <msup> <mi>q</mi> <mn>3</mn> </msup> <mo>+</mo> <mn>169</mn> <mi>q</mi> </mstyle> </math>.Suppose a revenue function is given by:

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:

R(q)=q3+169qR(q) = -q^3 + 169q

where qq is in thousands of units, and R(q)R(q) is in thousands of dollars.

(a) What is revenue when 4 thousand units are sold?

Substitute q=4q = 4 into the revenue function: R(4)=(4)3+169(4)R(4) = -(4)^3 + 169(4) R(4)=64+676=612 thousand dollars.R(4) = -64 + 676 = 612 \text{ thousand dollars.}

Answer (a): 612612 thousand dollars.


(b) Smallest qq for R(q)=361.152R(q) = 361.152 thousand dollars

We solve q3+169q=361.152-q^3 + 169q = 361.152. Rearrange: q3+169q361.152=0-q^3 + 169q - 361.152 = 0

This is a cubic equation. Solving it yields: q2.2 thousand units.q \approx 2.2 \text{ thousand units.}

Answer (b): 2.22.2 thousand units.


(c) Value of qq for maximum revenue

Revenue is maximized at the critical points, where the derivative of R(q)R(q) with respect to qq is zero: R(q)=3q2+169=0R'(q) = -3q^2 + 169 = 0 Solve for qq: 3q2+169=0    q2=1693    q=1693-3q^2 + 169 = 0 \implies q^2 = \frac{169}{3} \implies q = \sqrt{\frac{169}{3}} q7.5 thousand units (rounded to one decimal place).q \approx 7.5 \text{ thousand units (rounded to one decimal place).}

Answer (c): 7.57.5 thousand units.


Summary:

(a) 612612 thousand dollars
(b) 2.22.2 thousand units
(c) 7.57.5 thousand units

Would you like detailed step-by-step solutions or have any questions?

Here are 5 related questions for practice:

  1. How does increasing the unit price affect revenue for this function?
  2. Solve for qq when revenue is 500500 thousand dollars.
  3. What is the second derivative of the revenue function, and what does it indicate?
  4. If R(q)=q3+169q50R(q) = -q^3 + 169q - 50, what is the new revenue when q=4q = 4?
  5. 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