The top and bottom margins of a paster are each 6 cm and the side margins are each 4cm. If the area of printed material on the poster is fixed at 384cm^2, find the dimensinos of the poster with the smallest area.
I would like to know if my work is correct.
x = width of print
y = height of print
w = width of print
h = height of print
w = x + 8
h = y + 12
adding both sides of magins
A = wh
A = (x+8)(y + 12) xy = 384 y = 384/x
A = (x+8)(384/x + 12)
A = (x+8)([384 + 12x]/x)
A = (12x^2 + 480x + 3072)/x
domain:
x =/= 0 x > 0 cannot have negative width
(o,oo) open interval
differentiate:
A'(x) = (24x + 480)/-x^2
and this is where I get stuck, x is not able to be 0 yet I keep getting -20 to make the function = 0. Is the function actually a closed interval? I am actually having a hard time figuring out if something is an open or closed interval
Your A' is wrong. I expect the quotient rule got messy. Just make it a sum of powers of x:
A = 12x + 3072/x + 480
A' = 12 - 3072/x^2
A' = 0 when 12 = 3072/x^2
or, x = 16
so, y = 24
The printing is 16x24 and the poster is 24x36
Your work is on the right track, but there is a slight error in the differentiation step. Let me guide you through the process:
Given:
Top and bottom margins = 6cm each
Side margins = 4cm each
Area of printed material = 384cm^2
Let's assume:
Width of the print = x cm
Height of the print = y cm
To derive the dimensions of the poster with the smallest area, we need to formulate an equation for the area of the entire poster, including the margins.
Width of the entire poster = width of print + margins on both sides = x + 8 cm
Height of the entire poster = height of print + margins on top and bottom = y + 12 cm
Area of the entire poster = (width of the entire poster) * (height of the entire poster)
= (x + 8) * (y + 12)
Now, we have the equation for the area of the entire poster in terms of x and y. To find the dimensions of the poster with the smallest area, we can minimize this equation.
But instead of taking the derivative with respect to x, let's simplify the equation before taking the derivative.
Area of the entire poster = (x + 8) * (y + 12)
= xy + 8y + 12x + 96
Given that the area of the printed material is fixed at 384cm^2, we have:
xy = 384
Substituting this into our equation for the area of the entire poster:
Area of the entire poster = 384 + 8y + 12x + 96
= 480 + 8y + 12x
Now we have a simplified equation for the area of the entire poster.
To minimize this equation, let's take the partial derivatives with respect to both x and y:
dA/dx = 12
dA/dy = 8
Since both derivatives are constant, they are always zero. There is no need to set them equal to zero and solve for x or y. This means that the dimensions x and y of the printed material do not affect the minimum area, which is the fixed area of 384cm^2.
Therefore, we can conclude that the dimensions of the poster with the smallest area are x = width of the print = 16cm and y = height of the print = 24cm, since xy = 384.
In terms of intervals, since the dimensions of the print cannot be negative, the domain for the dimensions x and y is (0, infinity), which is an open interval.
I hope this clarifies the process and answer for you! Let me know if you have any further questions.