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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.)

Verified Answer :



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);

MrRoyal

The optimizing function of the printed area is \({P = (w-12) \times (h-20)}\), 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:

\(Area = hw\)

The area is given as 11760.

So, we have:

\(hw = 11760\)

Make w the subject

\(w = \frac{11760}h\)

When 10cm is removed from the height, and 6cm from the width.

The printed area (P) becomes

\({P = (w-12) \times (h-20)}\)

Recall that: \(w = \frac{11760}h}\)

So, we have:

\(P = (\frac{11760}{h}-12) \times (h-20)}\)

Expand

\(P = 11760 - \frac{235200}{h} - 12h + 240}\)

Differentiate

\(P' = 0+ \frac{235200}{h^2} - 12 + 0}\)

\(P' = \frac{235200}{h^2} - 12}\)

Set to 0

\(\frac{235200}{h^2} - 12 = 0}\)

Add 12 to both sides

\(\frac{235200}{h^2} = 12}\)

Multiply both sides by h^2

\(235200 = 12h^2}\)

Divide both sides by 12

\(19600 = h^2}\)

Take square roots of both sides

\({140 = h}\)

Rewrite as:

\({h =140 }\)

Recall that:

\(w = \frac{11760}h}\)

So, we have:

\(w = \frac{11760}{140}}\)

\(w = 84}\)

Hence, the dimensions that maximize the printed area is 84 by 140 cm

Read more about optimizing functions at:

https://brainly.com/question/6857653

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