Johnny is designing a rectangular poster to contain 15 in^2 of printing with a 5-in margin at the top and bottom and a 3-in margin at each side. What overall dimensions will minimize the amount of paper used?
Did you notice that in the "Related Questions" below I answered a similar question back in 2008.
All you have to do is change the numbers
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To minimize the amount of paper used, we need to find the overall dimensions of the rectangular poster that will contain 15 in^2 of printing.
Let's assume the width of the printing area is x inches. Since there is a 3-in margin on each side, the width of the overall poster will be x + 3 + 3 = x + 6 inches.
Similarly, let's assume the height of the printing area is y inches. Since there is a 5-in margin at the top and bottom, the height of the overall poster will be y + 5 + 5 = y + 10 inches.
The area of the printing region is given as 15 in^2, which corresponds to the width multiplied by the height:
x * y = 15 - (Equation 1)
The total area of the rectangular poster would be the width of the poster multiplied by the height:
(x + 6) * (y + 10) - (Equation 2)
To minimize the amount of paper used, we need to minimize the total area of the poster.
Now, let's substitute Equation 1 into Equation 2 to express the total area of the poster in terms of only one variable, x:
(x + 6) * (15/x + 10)
Simplifying this equation further will give us a quadratic expression, which we can minimize by finding the vertex of the quadratic equation.
To find the vertex, we can use the formula:
x = -b/2a
Where a = 15, and b = 40. Plugging in these values, we get:
x = -40 / (2 * 15)
x = -2.67
Since the width of the poster cannot be negative, we can discard this solution.
Therefore, the overall dimensions that will minimize the amount of paper used do not exist in this case.
To find the overall dimensions that minimize the amount of paper used, we need to consider the area of the rectangle and minimize it.
Let's denote the length of the poster's vertical side (including the printing and margins) as y and the length of the horizontal side as x.
From the given information, we know that the printed area of the poster is 15 square inches. Therefore, the area of the rectangle can be expressed as:
A = xy
Now let's calculate the total length of the vertical side of the poster:
Total vertical length = y + top margin + bottom margin
Total vertical length = y + 5 + 5
Total vertical length = y + 10
Similarly, let's calculate the total length of the horizontal side:
Total horizontal length = x + left margin + right margin
Total horizontal length = x + 3 + 3
Total horizontal length = x + 6
Now, the total area of the rectangle including the margins can be expressed as:
Total Area = (y + 10)(x + 6)
We want to minimize the total area, which means we need to find the critical points. Let's take the derivative of the total area equation with respect to y and set it equal to zero:
d(Total Area)/dy = 0
Expanding and simplifying the equation, we get:
0 = y + 10 - 15/(x + 6)
Now, let's solve this equation to find the critical point:
y + 10 = 15/(x + 6)
y(x + 6) + 10(x + 6) = 15
xy + 6y + 10x + 60 = 15
xy + 10x + 6y - 45 = 0
Now we have an equation in terms of two variables, x and y. To simplify this equation further, we need to substitute the given printed area:
xy = 15
Substituting xy = 15 into the equation, we get:
15 + 10x + 6y - 45 = 0
Now, we can find the value of x in terms of y:
10x + 6y = 30
10x = 30 - 6y
x = (30 - 6y)/10
x = (15 - 3y)/5
Now we can substitute this value of x back into the area equation and simplify it to get a quadratic equation:
Total Area = (y + 10)(x + 6)
Total Area = (y + 10)((15 - 3y)/5 + 6)
Expanding and simplifying the equation, we get:
Total Area = (y + 10)(15 - 3y + 30)
Total Area = (y + 10)(45 - 3y)
Now, we have a quadratic equation for the total area. To find the minimum value of the area, we can analyze the critical points.
To find the critical points, we take the derivative of the total area equation with respect to y:
d(Total Area)/dy = 0
Expanding and simplifying, we get:
0 = -6y^2 - 105y + 450
Dividing through by -3, we have:
0 = 2y^2 + 35y - 150
Now, we can solve this quadratic equation for y using factoring, completing the square, or the quadratic formula to find the critical values of y.
Once we find the critical values of y, we can substitute them back into the equation x = (15 - 3y)/5 to find the corresponding values of x.
Finally, substitute these values of x and y into the equation for the total area to find the minimum area.
The overall dimensions that minimize the amount of paper used will be the values of x and y obtained from this process.