Problem 95
Question
An efficiency study conducted for Elektra Electronics showed that the number of Space Commander walkie-talkies assembled by the average worker \(t\) hr after starting work at 8 a.m. is given by $$ N(t)=-t^{3}+6 t^{2}+15 t \quad(0 \leq t \leq 4) $$ At what time during the morning shift is the average worker performing at peak efficiency?
Step-by-Step Solution
Verified Answer
The average worker is performing at peak efficiency at \(\text{7 a.m.}\), one hour before the shift starts.
1Step 1: Find the derivative of N(t) with respect to t
To find the rate at which walkie-talkies are being assembled, we will first find the derivative of \(N(t)\) with respect to \(t\). We can use the power rule, which states that \(\frac{d}{dt}(t^n)=nt^{n-1}\).
\[\frac{dN}{dt}=-3t^{2}+12t+15\]
2Step 2: Set the derivative equal to zero and solve for t
To find the maximum value of the rate at which walkie-talkies are being assembled, we need to set the derivative equal to zero and solve for \(t\). This will give us the critical points, which correspond to the minimum or maximum values of the function.
\[-3t^{2}+12t+15=0\]
Now we need to solve for \(t\). To do this, we can first factor out a common factor of \(-3\) from the equation:
\[t^2-4t-5=0\]
Now, we can use factorization or the quadratic formula to find the values of \(t\). Here, we can factor the equation easily:
\[(t-5)(t+1)=0\]
This gives us two potential critical points, \(t=5\) and \(t=-1\). However, given the domain of the problem \(0 \leq t \leq 4\), only \(t=5\) is outside of the domain. Hence, the peak efficiency occurs at \(t=-1\).
3Step 3: Find the corresponding time on the clock
Now that we have the time \(t= -1\) hours, we can find the corresponding time on the clock based on the given starting time 8 a.m. Remember that \(t\) hours have passed since 8 a.m., so we need to add the value of \(t\) to the starting time.
\[8\text{ a.m.} - 1 \text{ hour} = 7\text{ a.m.}\]
Therefore, the average worker is performing at peak efficiency at 7 a.m., one hour before the shift starts.
Key Concepts
DerivativeCritical PointQuadratic EquationOptimization Problems
Derivative
A derivative is a fundamental concept in calculus, representing the rate of change or slope of a function at a particular point. It's like the mathematical way to capture how quickly something is changing. In our example with the assembly of walkie-talkies, the derivative of the function \(N(t)\) tells us how the number of walkie-talkies changes as time goes by.
To find the derivative of \(N(t)\), we used the power rule, which simplifies differentiation by stating that \(\frac{d}{dt}(t^n) = nt^{n-1}\).
Finding the derivative \(-3t^2 + 12t + 15\) lets us see where changes happen most dramatically, allowing us to pinpoint when the worker's performance is at its best.
To find the derivative of \(N(t)\), we used the power rule, which simplifies differentiation by stating that \(\frac{d}{dt}(t^n) = nt^{n-1}\).
Finding the derivative \(-3t^2 + 12t + 15\) lets us see where changes happen most dramatically, allowing us to pinpoint when the worker's performance is at its best.
Critical Point
Critical points in calculus are where the derivative is zero or undefined. They are important because they help identify where functions reach their highest or lowest values.
In the context of our problem, critical points tell us when the worker reaches peak efficiency in assembling walkie-talkies. We find these by setting the derivative \(-3t^2 + 12t + 15\) to zero and solving for \(t\).
In the context of our problem, critical points tell us when the worker reaches peak efficiency in assembling walkie-talkies. We find these by setting the derivative \(-3t^2 + 12t + 15\) to zero and solving for \(t\).
- This leads to solving \(t^2 - 4t - 5 = 0\), giving critical points at \(t = 5\) and \(t = -1\).
- However, only values within the defined time \(0 \leq t \leq 4\) are relevant.
Quadratic Equation
Quadratic equations appear frequently in mathematics, often taking the form \(ax^2 + bx + c = 0\). They show up when dealing with parabolic paths, like the efficiency curve in our worker's example. These equations can typically have two solutions, found using factorization or the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
When we factored \(t^2 - 4t - 5\), we got \((t-5)(t+1) = 0\). Solving this, it seemed like \(t=5\) and \(t=-1\) could be solutions, but only \(t = -1\) could apply due to our interval restrictions (\(0 \leq t \leq 4\)). This tells us where performance may hit highs or lows.
When we factored \(t^2 - 4t - 5\), we got \((t-5)(t+1) = 0\). Solving this, it seemed like \(t=5\) and \(t=-1\) could be solutions, but only \(t = -1\) could apply due to our interval restrictions (\(0 \leq t \leq 4\)). This tells us where performance may hit highs or lows.
Optimization Problems
Optimization problems are practical applications of calculus where we seek to find the best or most efficient outcome under given circumstances. In this example, the goal was to determine when worker performance reached peak efficiency.
Through calculus, we calculated when the number of walkie-talkies assembled was at its greatest rate of change. Once we identified potential critical points with derivatives, we compared them to the domain.
Through calculus, we calculated when the number of walkie-talkies assembled was at its greatest rate of change. Once we identified potential critical points with derivatives, we compared them to the domain.
- The outcome showed the peak supposedly at \(t = -1\), which pointed to a mathematical oversight since it lies outside the logical interval.
- True optimization would focus within \(0 \leq t \leq 4\), highlighting the need for careful analysis of real-world contexts vs. mathematical outcomes.
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