What dimensions will give the largest printed area? Following will be considered when answering the problem. a) Find the equation to maximized or minimized. b) Finding the solution c) Showing that your solution is an absolute maximized or minimized. A poster is to have an area of 180 in^2 with 1-inch margins at the bottom and sides and a 2-inch margin at the top. What dimensions will give the largest printed area?

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y-You should come up with the notations for the length and the width of poster: x the width and y the length.

The dimensions of printable area are as it follows: x-2 expresses the width and y - 3 the length.

You need to evaluate the area of poster:

`A =...

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y-You should come up with the notations for the length and the width of poster: x the width and y the length.

The dimensions of printable area are as it follows: x-2 expresses the width and y - 3 the length.

You need to evaluate the area of poster:

`A = x*y`

The problem provides the information that the area of the poster is of 180 in^2 such that:

`x*y = 180`

You should evaluate the printable area such that:

`A = (x-2)*(y-3)`

You need to express the area in terms of x only such that:

`A(x) = (x-2)*(180/x - 3)`

You need to open the brackets such that:

`A(x) = 180 - 3x - 360/x + 6`

a) You need to differentiate the function  A(x) with respect to x to find the equation that it is used in optimization such that:

`A'(x) = -3 + 360/x^2`

b) You need to solve the equation A'(x) = 0 such that:

`A'(x) =0 =gt -3 + 360/x^2 = 0 =gt -3x^2 + 360 = 0`

`x^2 - 120 = 0 =gt x^2 = 120 =gt x_1 = sqrt120 = 2sqrt30`

`x_2 = -2sqrt30`

Hence, the solutions to equation `A'(x) =0`  are `x_1 = 2sqrt30 ` and `x_2 = -2sqrt30` .

c) The function decreases over interval `(-2sqrt30 , 2sqrt30)`  and it increases over `(-oo,-2sqrt30)`  and `(2sqrt30 , oo)` , hence the function reaches its absolute maximum at `x=-2sqrt30`  and it reaches its absolute minimum at `x = 2sqrt30` .

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