A poster is to have an area of 160 in2 with 1 inch margins at the bottom and sides and a 7 inch margin at the top. What dimensions will give the largest printed area? (Give your answers correct to one decimal place.)
Well, to find the dimensions that will give the largest printed area, we need to consider the area of the poster minus the margins. Let's call the width of the poster "x" inches.
The height of the printed area will be x inches minus the 1-inch margins on the bottom and sides and the 7-inch margin at the top. So the height will be x - 1 - 1 - 7 = x - 9 inches.
Now, the printed area of the poster is the width multiplied by the height: x * (x - 9). We want to find the maximum value of this function.
To find the maximum, we can take the derivative of the function and set it equal to zero:
d(x * (x - 9))/dx = 2x - 9 = 0
Solving this equation, we get x = 4.5 inches.
Therefore, the dimensions that will give the largest printed area is a width of 4.5 inches and a height of (4.5 - 9) = -4.5 inches. Oops! That doesn't make sense.
Seems like we made a mistake somewhere. I apologize for any confusion caused. Let me try again with a different approach.
To find the dimensions that will give the largest printed area, we need to maximize the rectangular area while taking into account the given margins.
Let's assume the width of the poster is x inches. The height of the printed area would then be (160/x) square inches to have a total area of 160 square inches.
Given the margins, the actual width of the printed area would be (x - 2) inches (subtracting 1 inch from each side for the margins).
The actual height of the printed area would be [(160/x) - 7] inches (subtracting 7 inches for the margin at the top).
To maximize the printed area, we need to maximize the product of the width and the height.
So, the objective function (f(x)) is given by: f(x) = (x - 2) * [(160/x) - 7]
To find the dimensions that give the largest printed area, we need to find the maximum value of f(x).
To do this, we can take the derivative of f(x) with respect to x and set it equal to zero.
f'(x) = [(160/x) - 7] + (x - 2) * [-160/x^2]
Setting f'(x) = 0:
[(160/x) - 7] + (x - 2) * [-160/x^2] = 0
To simplify this equation, we can multiply through by x^2:
160 - 7x + (-160)(x - 2) = 0
160 - 7x - 160x + 320 = 0
-167x + 480 = 0
167x = 480
x = 2.874
Now we need to consider the constraints - the minimum value of x has to be greater than 3 (taking the 1-inch margin into account), and the maximum value of x can be 160 (assuming a minimum printed area of 1 square inch).
Since the minimum value of x is greater than 3, we can proceed.
Let's substitute this value of x into the dimension equations to find the corresponding height:
Actual width = (x - 2) = (2.874 - 2) = 0.874 inches
Actual height = [(160/x) - 7] = [(160/2.874) - 7] = 48.433 inches
So, the dimensions that will give the largest printed area are approximately 0.874 inches by 48.433 inches (correct to one decimal place).
To find the dimensions that give the largest printed area, we need to consider the margins and the total area. Let's start by calculating the available area for printing.
First, we need to determine the height available for printing. The total height of the poster is the sum of the top margin, the actual printed area, and the bottom margin.
Total Height = Top Margin + Printed Area + Bottom Margin
Total Height = 7 inches + Printed Area + 1 inch
Next, we need to determine the width available for printing. The total width of the poster is the sum of the left margin, the actual printed area, and the right margin.
Total Width = Left Margin + Printed Area + Right Margin
Total Width = 1 inch + Printed Area + 1 inch
Given that the total area of the poster is 160 in^2 and the height and width are related by the formula: Total Area = Total Height * Total Width, we can set up the equation:
160 in^2 = (7 in + Printed Area + 1 in) * (1 in + Printed Area + 1 in)
Simplifying this equation:
160 in^2 = (Printed Area + 8 in) * (Printed Area + 2 in)
Expanding the equation:
160 in^2 = Printed Area^2 + 10 in * Printed Area + 16 in^2
Rearranging the equation:
Printed Area^2 + 10 in * Printed Area + 16 in^2 - 160 in^2 = 0
Simplifying further:
Printed Area^2 + 10 in * Printed Area - 144 in^2 = 0
Now, we have a quadratic equation where the printed area is the variable. We can solve this quadratic equation to find the dimensions that give the largest printed area.