The top and bottom margins of a poster are each 12 cm and the side margins are each 8 cm. If the area of printed material on the poster is fixed at 1536 cm^2, find the dimensions of the poster with the smallest area.
48
Well, it seems like this poster wants to be a minimalist. With those small margins, it just wants to get straight to the point!
Let's call the width of the printed material "w" and the height "h". So we have two unknowns here, just like in a math mystery!
Now, we know that the top and bottom margins are 12 cm each, so the total height of the poster will be h + 2(12) = h + 24. Similarly, the total width of the poster will be w + 2(8) = w + 16.
The area of the printed material is fixed at 1536 cm^2, so we can set up an equation: (w + 16) * (h + 24) = 1536.
Now, let's delve into the world of optimization to find the smallest area for our minimalist poster. To do this, we need to minimize w * h while keeping the above equation in mind.
Without further ado, let's solve this puzzle!
To find the dimensions of the poster with the smallest area, we can use the concept of optimization by setting up a mathematical model and then differentiating to find the critical points.
Let's assume the length of the poster is L and the width is W.
Given that the top and bottom margins are each 12 cm, and the side margins are each 8 cm, we can determine the dimensions of the printed material on the poster. The printed area will be the total area of the poster minus the margins.
Therefore, the dimensions of the printed material will be (L - 2 × 8) cm for the length and (W - 2 × 12) cm for the width.
Now, from the given information, we know that the area of the printed material is fixed at 1536 cm^2. So we have the equation:
(L - 2 × 8) × (W - 2 × 12) = 1536
Simplifying the equation further:
(L - 16) × (W - 24) = 1536
Expanding the equation:
LW - 24L - 16W + 384 = 1536
Combining like terms:
LW - 24L - 16W + 384 - 1536 = 0
LW - 24L - 16W - 1152 = 0
Now, the goal is to find the critical points, so we need to determine the partial derivatives.
Taking the partial derivative of the equation with respect to L:
∂/∂L (LW - 24L - 16W - 1152) = W - 24
Taking the partial derivative of the equation with respect to W:
∂/∂W (LW - 24L - 16W - 1152) = L - 16
To find the critical points, set both partial derivatives equal to zero and solve the system of equations:
W - 24 = 0
L - 16 = 0
Solving these two equations, we find:
W = 24
L = 16
Therefore, the length of the poster is 16 cm and the width is 24 cm.
Let the width of the image (excluding margins) be x, then the height is 1536-x.
The total area of the poster,
A(x)=(x+2*8)(1536/x+2*12)
Differentiate with respect to x, equate to zero and solve for x. I get x=32.
Is there anything that you don't follow?
Width=48
height=32