The top and bottom margins of a poster are 2 cm and the side margins are each 4 cm. If the area of printed material on the poster is fixed at 386 square centimeters, find the dimensions of the poster with the smallest area.
To find the dimensions of the poster with the smallest area, we can use the given information and apply the formula for area.
Let's suppose the width of the printed material is x cm.
Since the top and bottom margins are both 2 cm, the total vertical dimension of the poster would be:
2 cm (top margin) + x cm (printed material) + 2 cm (bottom margin) = x + 4 cm
Similarly, since the side margins are each 4 cm, the total horizontal dimension of the poster would be:
4 cm (left margin) + x cm (printed material) + 4 cm (right margin) = x + 8 cm
The area of the poster can then be calculated by multiplying the two dimensions:
Area = (x + 4 cm) * (x + 8 cm)
Given that the area of the printed material on the poster is fixed at 386 square centimeters, we can set up the following equation:
(x + 4 cm) * (x + 8 cm) = 386 cm²
Now, let's solve this equation to find the value of x and determine the dimensions of the poster with the smallest area.
Expanding the equation: x² + 12x + 32 = 386
Rearranging the equation: x² + 12x + 32 - 386 = 0
Simplifying: x² + 12x - 354 = 0
Now, we can solve this quadratic equation. There are a few methods to solve quadratic equations, such as factoring, completing the square, or using the quadratic formula. Let's use the quadratic formula since it can be applied to any quadratic equation.
The quadratic formula states that for an equation of the form ax² + bx + c = 0, the value of x can be found using the formula:
x = (-b ± √(b² - 4ac)) / (2a)
Applying the quadratic formula to our equation (x² + 12x - 354 = 0), we have:
x = (-12 ± √(12² - 4 * 1 * -354)) / (2 * 1)
Simplifying further: x = (-12 ± √(144 + 1416)) / 2
x = (-12 ± √(1560)) / 2
Now, let's calculate the two possible values of x using a calculator:
x₁ = (-12 + √1560) / 2 ≈ 12.22 cm
x₂ = (-12 - √1560) / 2 ≈ -24.22 cm
Since the dimensions cannot be negative, we ignore the negative value of x.
Therefore, the width of the printed material on the poster is approximately 12.22 cm.
To find the dimensions of the poster, we can substitute this value back into the earlier calculations:
Vertical dimension: x + 4 cm = 12.22 cm + 4 cm = 16.22 cm
Horizontal dimension: x + 8 cm = 12.22 cm + 8 cm = 20.22 cm
Therefore, the dimensions of the poster with the smallest area are approximately 12.22 cm by 16.22 cm.
To find the dimensions of the poster with the smallest area, we need to minimize the area of the printed material.
Let's assume the length of the poster is L and the width is W.
The length of the printed material will be L - 2 cm (since there are 2 cm top and bottom margins) and the width will be W - 2 * 4 cm = W - 8 cm (since there are 4 cm side margins on each side).
The area of the printed material is given by multiplying the length and width:
Area of printed material = (L - 2 cm) * (W - 8 cm)
We are given that the area of the printed material is fixed at 386 square centimeters, so we can write the equation:
(L - 2 cm) * (W - 8 cm) = 386 square centimeters
Expanding the equation:
LW - 8L - 2W + 16 = 386
Now, we want to find the dimensions that will minimize the area of the printed material. In this case, we need to find the minimum value of the function LW - 8L - 2W.
To find the minimum value, we can take the partial derivatives of the function with respect to L and W, and set them equal to zero:
d/dL (LW - 8L - 2W) = 0
d/dW (LW - 8L - 2W) = 0
Differentiating the function with respect to L:
W - 8 = 0
W = 8
Similarly, differentiating the function with respect to W:
L - 2 = 0
L = 2
So, the dimensions of the poster with the smallest area are L = 2 cm and W = 8 cm.
If the printed area has width x and height y,
xy = 386
The total area is thus
a = (x+8)(y+4) = (x+8)(386/x + 4)
= 4x + 418 + 3088/x
max area is where da/dx=0
da/dx = 4 - 3088/x^2
so, da/dx = 0 when x = 2√193
Now just evaluate y.