A poster of area 11,760 cm2 has blank margins of width 10 cm on the top and bottom and 6 cm on the sides. Find the dimensions that maximize the printed area. (Let w be the width of the poster, and let h be the height.)

w=width of the poster.

h=hegith of the poster.

Area of the poster = wh

Then: wh=11760 ⇒w = (11760/h)

let:

P (w, h) = printed area

P (w, h) = (w-12) (h-20)

P (h) = (11760/h - 12) (h-20)

We have a problem of maximums and minimums.

1) we compute the first derivative of this function:

P (h) = (11760/h - 12) (h-20)

P' (h) = (-11760/h²) (h-20) + (11760/h-12) =

(-11760h+235200) / h² + (11760-12h) / h=

(-11760h+235200+11760h-12h²) / h²=

(235200-12h²) / h²

Therefore:

P' (h) = (235200-12h²) / h²

2) we find the values of "h" when P (h) = 0.

h≠0

(235200-12h²) / h²=0

(235200-12h²) = 0 (h²)

235200-12h²=0

-12h²=-235200

h²=-235200/-12

h²=19600

h=√19600

h=140

3) we have to find the second derivative.

P' (h) = (235200-12h²) / h²

P'' (h) = [-24h (h²) - 2h (235200-12h²) ] / h⁴

P'' (h) = [-24h²-2 (235200-12h²) ]/h³

P'' (h) = (-24 h²-470400+24h²) / h³

P'' (h) = - 470400 / h³

P'' (140) = - 470400 / (140³) <0, therefore we have a maximum at h=140.

4) we find out the width

w = (11760/h)

w=11760/140

w=84

Answer: the dimensions that maximize the prited area would be:

84 x 140 (cm);