Problem 26 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

In Exercises 25 through 28, find the absolute maximum and the absolute minimum values (if any) of the given function on the specified interval. $$ f(t)=-3 t^{4}+8 t^{3}-10 ; 0 \leq t \leq 3 $$

Step-by-Step Solution

Answer

Absolute maximum value is 30 at \[t = 2\]; absolute minimum value is -37 at \[t = 3\].

Step 1: Identify the function and interval

The given function is \[f(t) = -3t^4 + 8t^3 - 10\] and the interval is \[0 \leq t \leq 3\].

Step 2: Find the derivative of the function

To find the critical points, compute the first derivative \[f'(t)\]: \[f'(t) = \frac{d}{dt}(-3t^4 + 8t^3 - 10) = -12t^3 + 24t^2\].

Step 3: Solve for critical points

Set the first derivative equal to zero and solve for \[t\]: \[-12t^3 + 24t^2 = 0\] Factor out the common term \[t^2\]: \[-12t^2(t - 2) = 0\] This gives critical points \[t = 0\] and \[t = 2\].

Step 4: Evaluate the function at the critical points and endpoints

Evaluate the function \[f(t)\] at the critical points \[t = 0\] and \[t = 2\], and at the endpoints \[t = 0\] and \[t = 3\]: \[f(0) = -3(0)^4 + 8(0)^3 - 10 = -10\]\[f(2) = -3(2)^4 + 8(2)^3 - 10 = -24 + 64 - 10 = 30\]\[f(3) = -3(3)^4 + 8(3)^3 - 10 = -243 + 216 - 10 = -37\].

Step 5: Determine the absolute maximum and absolute minimum values

Compare the values: \[f(0) = -10\],\[f(2) = 30\],\[f(3) = -37\]. The absolute maximum value is \[30\] at \[t = 2\], and the absolute minimum value is \[-37\] at \[t = 3\].

Key Concepts

Critical Points

To find the absolute maximum and minimum of a function, identifying the critical points is crucial. These points occur where the function's derivative is zero or does not exist. For our function \[ f(t) = -3t^4 + 8t^3 - 10 \], we begin by computing its derivative: \[ f'(t) = -12t^3 + 24t^2 \]. Next, we set this derivative equal to zero to find the critical points: \[ -12t^3 + 24t^2 = 0 \]. Factoring out the common term gives us: \[ -12t^2(t - 2) = 0 \]. From here, we see that our critical points are \[ t = 0 \] and \[ t = 2 \]. These values will be essential when evaluating our function at critical points and endpoints within the given interval.

Derivative of a Function

In solving for the extreme values of a function, derivatives play a fundamental role. The derivative represents the rate at which the function changes, providing insights into the function's slope at any given point. For our function \[ f(t) = -3t^4 + 8t^3 - 10 \], the first step in finding critical points is computing the derivative: \[ f'(t) = -12t^3 + 24t^2 \]. We derived this by applying the power rule. For a general term \[ at^n \], the power rule states that its derivative is \[ an t^{n-1} \]. This leads us to: \[ -3 \times 4 t^{4-1} + 8 \times 3 t^{3-1} = -12t^3 + 24t^2 \]. By understanding how derivatives work, you can confidently find and interpret critical points and other key features of a function.

Interval Evaluation

Analyzing a function within a specified interval helps in identifying the absolute maximum and minimum values. For the given problem, our interval is \[ 0 \leq t \leq 3 \]. This interval ensures we only focus on the specified domain of \[ f(t) \]. To determine the extrema, we must evaluate \[ f(t) \] at both the critical points and the endpoints of the interval. Let's evaluate the function: \[ f(0) = -10 \] (left endpoint), ü \[ f(2) = 30 \] (critical point), \[ f(3) = -37 \] (right endpoint). Among these values, identify the highest and lowest points. Here, \[ f(2) = 30 \] is the maximum and \[ f(3) = -37 \] is the minimum. Performing interval evaluations helps pinpoint these absolute values precisely.

Calculus Optimization

Optimization in calculus involves finding the maximum or minimum value of a function. In our problem, we aim to determine where \[ f(t) \] reaches its highest and lowest values within the given interval. First, computing the derivative and finding the critical points helps us locate where the function's slope is zero or undefined. Next, evaluating the function at these critical points and the interval endpoints reveals the function's extreme values. For \[ f(t) = -3t^4 + 8t^3 - 10 \], within \[ 0 \leq t \leq 3 \], our evaluations showed \[ f(0) = -10 \], \[ f(2) = 30 \], and \[ f(3) = -37 \]. Thus, the absolute maximum is \[ 30 \] at \[ t = 2 \], while the absolute minimum is \[ -37 \] at \[ t = 3 \]. Through these steps, calculus optimization effectively helps you determine the important critical values of any function in any interval.