Problem 83 - University Calculus: Early Transcendentals

Question

The function $$V(x)=x(10-2 x)(16-2 x), \quad 0

Step-by-Step Solution

Answer

Maximum volume occurs at \( x = 2 \) with 144 cubic meters; zero volume occurs near \( x = 0 \) and \( x = 5 \).

Step 1: Identify the Function

The given function is \( V(x) = x(10 - 2x)(16 - 2x) \). This function models the volume of a box, where \( x \) is a dimension of the box between 0 and 5.

Step 2: Expand the Function

First, expand the product in the volume function. Simplify it to find a polynomial expression for easier differentiation. \[V(x) = x((10 - 2x)(16 - 2x))\]First, expand \((10 - 2x)(16 - 2x)\):\[= 160 - 20x - 32x + 4x^2 = 160 - 52x + 4x^2\]Multiply this by \( x \):\[V(x) = x(160 - 52x + 4x^2) = 4x^3 - 52x^2 + 160x\]

Step 3: Find the Derivative of the Function

To find the critical points of \( V(x) \), we need to take its derivative with respect to \( x \) and set it to zero:\[V'(x) = \frac{d}{dx}(4x^3 - 52x^2 + 160x) = 12x^2 - 104x + 160\]

Step 4: Solve for Critical Points

Set \( V'(x) = 0 \) to find critical points:\[12x^2 - 104x + 160 = 0\]Simplify and solve this quadratic equation using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \( a = 12 \), \( b = -104 \), and \( c = 160 \):\[x = \frac{104 \pm \sqrt{(-104)^2 - 4 \times 12 \times 160}}{24}\]\[x = \frac{104 \pm \sqrt{10816 - 7680}}{24}\]\[x = \frac{104 \pm \sqrt{3136}}{24}\]\[x = \frac{104 \pm 56}{24}\]The solutions are \( x = \frac{160}{24} = \frac{20}{3} \approx 6.67 \) and \( x = 2 \). Since \( 0 < x < 5 \), only \( x = 2 \) is valid in this interval.

Step 5: Know the Extreme Values

Check the endpoints and the critical point \( x = 2 \) to find extreme values of the function within the interval. Compute \( V(x) \) at these points:- \( V(0^+) = 0 \, m^3 \) because volume approaches 0 as \( x \) approaches 0.- \( V(5^-) = 0 \, m^3 \) since the width term \((10 - 2x)\) becomes zero.- \( V(2) = 4(2)^3 - 52(2)^2 + 160(2) = 32 - 208 + 320 = 144 \, m^3 \).Thus, the maximum volume is \( 144 \), occurring at \( x = 2 \).

Step 6: Interpretation of Extreme Values

The extreme values indicate the possible volume variations within the given interval. A volume of 0 at \( x = 0 \) and \( x = 5 \) is logical because one or more box dimensions become zero, reducing the volume to zero. The maximum volume \( 144 \, m^3 \) occurs at \( x = 2 \); thus, choosing this dimension for the box yields maximum volumetric efficiency.

Key Concepts

Volume of a Box

Understanding how the volume of a box is calculated is crucial when working with geometric problems. Here, we have a function that represents the volume of a box:

1. The given function is \( V(x) = x(10 - 2x)(16 - 2x) \). This means that the volume is dependent on the variable \( x \), which is one dimension of the box, affecting the other dimensions as they reduce by \( 2x \).
2. To find the fullest capacity of the box within a specified range of \( x \), we must evaluate how \( x \) changes the overall volume, especially by expanding the expression and exploring its polynomial form.

Knowing how each dimension shrinks as \( x \) increases, students can visualize this as a box being cut from a piece of material, where the cuts create flaps to form the box. Calculating such a volume variation helps in understanding how "real-life" cutting problems work to maximize storage space.

Critical Points

In mathematical analysis, finding the critical points of a function tells us where the maximum and minimum values might occur. Here’s a simple breakdown of what critical points mean:

• Critical points occur where the derivative of a function is zero or undefined. In our instance, we got the derivative \( V'(x) \) and set it equal to zero to locate these points: \( 12x^2 - 104x + 160 = 0 \).
• By solving the above quadratic equation with the quadratic formula, we found the values of \( x \) that make the derivative zero, which are possible critical points at \( x = 2 \) within the interval \( 0 < x < 5 \).

Once critical points are identified, we not only have information about where optimal values might lie, but also how the function behaves as \( x \) moves through the range. Remember, especially when considering real-world applications, boundaries play a crucial role in determining feasible solutions.

Polynomial Differentiation

Polynomial differentiation is a primary tool in calculus used to find critical points, like those we explored above. This technique involves the differentiation of polynomial expressions to understand the rate of change of quantities.

• To differentiate a polynomial like \( V(x) = 4x^3 - 52x^2 + 160x \), you apply the power rule, which states: for any term \( ax^n \), the derivative is \( nax^{n-1} \).
• This led us to the result \( V'(x) = 12x^2 - 104x + 160 \), where each term of the polynomial \( 4x^3 \), \( -52x^2 \), and \( 160x \) was differentiated with respect to \( x \).

The outcome of differentiating polynomials is powerful as it provides insight into how changing conditions affect outcomes, like maximizing a box's volume by understanding how each of its sides adjust via the changing parameter \( x \). This foundational method is vital for any optimization problem in calculus.