The total profit P(x), in thousands of Naira, obtained from producing and selling x hundre… — JAMB UTME

The total profit $P(x)$ , in thousands of Naira, obtained from producing and selling $x$ hundred units of a commodity is modeled by the function $P(x) = -x^3 + 6x^2 + 15x - 8$ , where $x \ge 0$ . What is the maximum profit achievable?

Answer: 92 thousand Naira

Answer

The maximum profit achievable is 92 thousand Naira.

To find the maximum profit, we find the stationary points of

P(x) = -x^3 + 6x^2 + 15x - 8

by taking the derivative

P'(x) = -3x^2 + 12x + 15

and setting it to 0. Solving

-3(x-5)(x+1) = 0

with

x \ge 0

yields

x = 5

. Testing the second derivative gives

P''(5) = -18 < 0

, confirming

x = 5

is a maximum. Substituting

x = 5

into

P(x)

yields

P(5) = 92

Step-by-Step Solution

Differentiate the profit function

P(x)

with respect to

x

P'(x) = -3x^2 + 12x + 15

Stationary points occur where the rate of change of profit (the derivative) is equal to zero.

Set

P'(x) = 0

and solve for

x

-3(x^2 - 4x - 5) = 0 \implies (x - 5)(x + 1) = 0

, giving critical values

x = 5

and

x = -1

Factoring the quadratic equation yields the critical values of production level.

Filter critical values based on physical domain constraints

x = 5

(reject

x = -1

since production

x \ge 0

)

Production quantities cannot be negative in physical real-life contexts.

Perform the second derivative test to confirm the nature of the stationary point

P''(x) = -6x + 12 \implies P''(5) = -6(5) + 12 = -18

Since

P''(5) < 0

, the function achieves a local maximum at

x = 5

Substitute

x = 5

back into original function

P(x)

P(5) = -(5)^3 + 6(5)^2 + 15(5) - 8 = -125 + 150 + 75 - 8 = 92

Evaluating

P(5)

gives the maximum total profit value.

Key Concept

Optimization and Maximum Values using First and Second Derivatives

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