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Calculus IChapter 4: Application of DerivativesSection 4-4

Finding Absolute Extrema — Question 7

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Question 7

Problem:

A company models its daily profit (in hundreds of dollars) as a function of the number of units produced, using the function: P(x)=−x3+6x2+15P(x) = -x^3 + 6x^2 + 15 where xx is the number of units (in hundreds) produced per day, and x∈[0,5]x \in [0, 5].

  • (a) Find all critical points of P(x)P(x) on the interval.

  • (b) Determine the absolute maximum and minimum profit on the interval [0,5][0, 5].

Original worksheet page 1: question and worked solution for 4-4-007
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Question 7 - Solution

We are given: P(x)=−x3+6x2+15P(x) = -x^3 + 6x^2 + 15

(a) Find critical points:

Differentiate: P′(x)=−3x2+12x=−3x(x−4)P'(x) = -3x^2 + 12x = -3x(x - 4)

Set derivative equal to zero: −3x(x−4)=0⇒x=0,x=4-3x(x - 4) = 0 \Rightarrow x = 0, \, x = 4

These are critical points in the interval [0,5][0, 5].

(b) Evaluate endpoints and critical points:

P(0)=−0+0+15=15P(0) = -0 + 0 + 15 = 15 P(4)=−64+96+15=47P(4) = -64 + 96 + 15 = 47 P(5)=−125+150+15=40P(5) = -125 + 150 + 15 = 40

Conclusion:

  • Absolute Maximum: P(4)=47\boxed{P(4) = 47}

  • Absolute Minimum: P(0)=15\boxed{P(0) = 15}

Graph of P(x)=−x3+6x2+15P(x) = -x^3 + 6x^2 + 15 on [0,5][0, 5]:

See the diagram in the original worksheet below.

Original worksheet page 2: question and worked solution for 4-4-007

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