Aposter 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

the dimensions that maximize the prited area would be:
 84 x 140 (cm);