Hi, how do I go about answering this question?
"An efficiency study of the morning shift at a certain factory indicates that an average worker who is on the job at 8:00 A.M. will have assembled f(x)= -x^3+6x^2+15x units x hours later. The study indicates further that after a 15-minute coffee break the worker can assemble g(x) = (-1/3)x^3+x^2+23x units in x hours.
Determine the time between 8:00 A.M. and noon at which a 15-minute coffee break should be scheduled so that the worker will assemble the maximum number of units by lunch time at 12:15 P.M."
Thank you!
"An efficiency study of the morning shift at a certain factory indicates that an average worker who is on the job at 8:00 A.M. will have assembled f(x)= -x^3+6x^2+15x units x hours later. The study indicates further that after a 15-minute coffee break the worker can assemble g(x) = (-1/3)x^3+x^2+23x units in x hours.
Determine the time between 8:00 A.M. and noon at which a 15-minute coffee break should be scheduled so that the worker will assemble the maximum number of units by lunch time at 12:15 P.M."
Thank you!
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Let h(x) denote the number of units assembled if the coffee break is taken x hours after 8am. Then
h(x) = f(x) + g(4-x)
= -x³ + 6x² + 15x - (1/3)(4 - x)³ + (4 - x)² + 23(4 - x)
for 0 ≤ x ≤ 4
so we are looking for the absolute max of h on [0, 4]. The conditions of the Extreme Value Theorem are met, and so the absolute max (and min) of h occur at an endpoint of the interval or at a critical point in the interval. So we need dh/dx.
Though you don't have to, I would recommend expanding out the above expression before differentiating . I get
h(x) = (-2/3)x³ + 3x² + (260/3)
Then
dh/dx = -2x² + 6x = 2x(3 - x)
dh/dx = 0 ⇒ x = 0 or x = 3
Compute h(0), h(3), and h(4). Pick the x-value that gives the largest h.
h(x) = f(x) + g(4-x)
= -x³ + 6x² + 15x - (1/3)(4 - x)³ + (4 - x)² + 23(4 - x)
for 0 ≤ x ≤ 4
so we are looking for the absolute max of h on [0, 4]. The conditions of the Extreme Value Theorem are met, and so the absolute max (and min) of h occur at an endpoint of the interval or at a critical point in the interval. So we need dh/dx.
Though you don't have to, I would recommend expanding out the above expression before differentiating . I get
h(x) = (-2/3)x³ + 3x² + (260/3)
Then
dh/dx = -2x² + 6x = 2x(3 - x)
dh/dx = 0 ⇒ x = 0 or x = 3
Compute h(0), h(3), and h(4). Pick the x-value that gives the largest h.