Question

Question #5cf1a

Answer 1

The smallest area is `864 \ cm^2` which occurs when the dimensions are `36 \ cm xx 24 \ cm`

Explanation:

Let us set up the following variables:

`{(x, "Width of poster (cm)"), (y, "Height of poster (cm)"), (A, "Area of poster ("cm^3")") :}`

Then the dimensions of the printed matter are:

`{("Width", =x-6-6,=x-12), ("Height", =y-4-4,=y-8), ( :. " Area",=384 ,=(x-12)(y-8)) :}`

So we have;

`384 =(x-12)(y-8)`
`:. y-8 = 384/(x-12)`
`:. y = 8+384/(x-12)`
`" " = (8(x-12)+384)/(x-12)`
`" " = (8x-96+384)/(x-12)`
`" " = (8x+288)/(x-12) \ \ \ \ \ ..... (star)`

And the total area of the poster is given by:

`A = xy`
`\ \ \ = (x)((8x+288)/(x-12)) \ \ \` (using (`star`))
`\ \ \ = (8x^2+288x)/(x-12)`

We want to minimize (hopefully) by finding `(dA)/dx`, which we get by applying the product rule:

`(dA)/dx = { (x-12)(16x+288) - (1)(8x^2+288x) } / (x-12)^2`
`" " = { 16x^2+288x-192x+3456-8x^2-288x } / (x-12)^2`
`" " = { 8x^2-192x+3456 } / (x-12)^2`

At a min or max `(dA)/dx=0`

`:. { 8x^2-192x+3456 } / (x-12)^2 = 0`
`:. 8x^2-192x-3456 = 0`
`:. 8(x^2-24x-432)=0`
`:. x^2-24x-432 = 0`
`:. (x+12)(x-36) = 0`

This equation leads to the two solutions:

`x= -12`, or `x=36`

Obviously `x>0`, so we can eliminate `x=-12`, leaving `x=36` as the only valid solution. With this value of `x` we have:

`y = (288+288)/(36-12) \ \ \ \ \` (using (`star`))
`\ \ = 576/24`
`\ \ = 24`

With these dimensions we have:

`A = 36*24`
`\ \ \ = 864`

We should check that this value leads to a minimum (rather than a maximum) area, which we confirm graphically:

graph{(8x^2+288x)/(x-12) [-2, 50, -2000, 2000]}