Question

Designing a Poster You are designing a rectangular poster to contain 50 in \(^{2}\) of printing with a 4 -in. margin at the top and bottom and a 2-in. margin at each side. What overall dimensions will minimize the amount of paper used?

Short answer

The overall dimensions of the poster that would minimize paper usage are 9 in. in width and 18 in. in height.

Step-by-step solution

Define the Variables

Let's denote the width of the printed area by \( x \) and the height by \( y \). Then, the width of the whole poster will be \( x+2*2 \) because there's a 2 in. margin at each side, and the height would be \( y+2*4 \) due to the 4 in. margin at the top and bottom.

Formulate the Equations

The area for printing is 50 in^{2}, which gives us the equation \( x*y = 50 \). The total area of the poster can be represented as \( (x+4)*(y+8) \).

Express One Variable in Terms of Another

From the equation \( x*y = 50 \), we can express \( y \) as \( y = 50/x \).

Substitute y in the Total Area Equation

Substitute \( y = 50/x \) into the total area equation to get a function depending on only one variable. This gives \( (x+4)*((50/x)+8) \).

Simplify the Equation

After simplifying, we obtain the equation \( ((50/x)+8)*(x+4) = 50/x * 4+x*8+4*8 \).

Calculate the Derivative

Now we have to find the derivative to minimize the area. The derivative of the function is \( \frac{dA}{dx} = 200/x^2 - 8 \).

Set the Derivative Equal to Zero

The critical points are obtained when the derivative is equal to zero, which gives us \( 200/x^2 - 8 = 0 \). Solving this equation gives \( x = 5 \).

Substitute \( x = 5 \) into Equations

Substitute \( x = 5 \) back into the initial equation \( y = 50/x \) to find \( y = 10 \). Adding the margins, the dimensions of the poster that would minimize the paper used are \( 5+4 = 9 \) in width and \( 10+8 = 18 \) in height.

Key concepts

Calculus Problem Solving

Calculus is often seen as a tool for engineers and scientists, but it's just as valuable when it comes to problem-solving in everyday contexts, such as designing a poster efficiently. In the given exercise, the student was asked to determine the dimensions of a poster that would minimize the amount of paper used while containing a specific printing area. This task is a classic example of an optimization problem.

When approaching such calculus problems, the first crucial step is defining the variables. The chosen variables represent quantities we want to control or those that affect the outcome—here, the width and height of the printed area. With the variables in hand, we then build equations representing the constraints and goals of the problem. Here, we needed to respect the printing area and minimize the total paper area used.

Next comes expressing one variable in terms of another to reduce the problem to a single-variable scenario. This manipulation is where calculus starts to shine as simplifying the equations can often lead to finding a function that we can optimize using the power of derivatives, as we will see in our next section.

Derivative Applications

With our problem simplified to a single variable, we apply the concept of derivatives to find the optimal solution. Derivatives are powerful tools in calculus that provide us with information about the rate at which a function is changing. In the context of optimization, we're interested in finding where this rate of change is zero—which is to say, where the function has a critical point and is no longer increasing or decreasing.

In our poster design problem, after setting up a function that models the total area in terms of one variable, the derivative gives us a formula to calculate how this area changes as the dimensions change. By setting this derivative equal to zero and solving for the variable, we pinpoint the dimensions where the change in area is neither increasing nor decreasing—a strong candidate for our minimum paper usage scenario.

However, just finding a critical point is not sufficient. We must also check whether this point is truly a minimum (and not a maximum or a point of inflection) by either using the second derivative test or by analyzing the sign of the derivative around the critical point. In this case, after finding the critical point where the derivative equals zero, we deduce that this indeed corresponds to the minimum area needed for the paper.

Minimizing Surface Area

Minimizing the surface area in mathematical terms means we are searching for the smallest possible value that the surface area can take under given constraints. This is done by establishing an equation for the surface area with respect to the important variables, and then finding where this area is at its minimum value. The process generally involves differentiation and setting a derivative equal to zero, as critical points indicate where a function may have local minima or maxima.

For the case at hand, minimizing the surface area meant optimizing the dimensions of printed and margin areas on a poster. After calculating the derivative of the area function and finding the critical point, we located the dimensions that minimize the surface area of the poster. Notably, effectiveness in minimizing comes from the close linkage between the algebraic manipulation of the area equation and the strategic use of calculus through differentiation.

Understanding these fundamental concepts and how they intertwine is paramount for students tackling similar calculus problems. By seeing the relationships between the practical task of designing a poster and the mathematical concepts used to optimize the process, students gain insight into the real-world applications of calculus beyond theoretical exercises.