Respuesta :
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);
The optimizing function of the printed area is [tex]{P = (w-12) \times (h-20)}[/tex], while dimensions that maximize the printed area is 84 by 140 cm
Let the dimension of the poster be w and h, where:
- w represents the width.
- h represents the height.
So, the area of the poster is:
[tex]Area = hw[/tex]
The area is given as 11760.
So, we have:
[tex]hw = 11760[/tex]
Make w the subject
[tex]w = \frac{11760}h[/tex]
When 10cm is removed from the height, and 6cm from the width.
The printed area (P) becomes
[tex]{P = (w-12) \times (h-20)}[/tex]
Recall that: [tex]w = \frac{11760}h}[/tex]
So, we have:
[tex]P = (\frac{11760}{h}-12) \times (h-20)}[/tex]
Expand
[tex]P = 11760 - \frac{235200}{h} - 12h + 240}[/tex]
Differentiate
[tex]P' = 0+ \frac{235200}{h^2} - 12 + 0}[/tex]
[tex]P' = \frac{235200}{h^2} - 12}[/tex]
Set to 0
[tex]\frac{235200}{h^2} - 12 = 0}[/tex]
Add 12 to both sides
[tex]\frac{235200}{h^2} = 12}[/tex]
Multiply both sides by h^2
[tex]235200 = 12h^2}[/tex]
Divide both sides by 12
[tex]19600 = h^2}[/tex]
Take square roots of both sides
[tex]{140 = h}[/tex]
Rewrite as:
[tex]{h =140 }[/tex]
Recall that:
[tex]w = \frac{11760}h}[/tex]
So, we have:
[tex]w = \frac{11760}{140}}[/tex]
[tex]w = 84}[/tex]
Hence, the dimensions that maximize the printed area is 84 by 140 cm
Read more about optimizing functions at:
https://brainly.com/question/6857653
