Question

You are designing a rectangular poster to contain 7575 in2 of printing with a 33​-in margin at the top and bottom and a 11​-in margin at each side. What overall dimensions will minimize the amount of paper​ used?

Answer 1

Answer:

The dimensions of the rectangular poster is 15 in by 5 in.

Step-by-step explanation:

Given that, the area of the rectangular poster is 75 in².

Let the length of the rectangular poster be x and the width of the rectangular poster be y.

The area of the poster = xy in².

$\therefore xy=75$

$\Rightarrow y=\frac{75}{x}$....(1)

1 in margin at each sides and 3 in margin at top and bottom.

Then the length of printing space is= (x-2.3) in

                                                           =(x-6) in

The width of printing space is = (y-2.1) in

                                                  =(y-2) in

The area of the printing space is A =(x-6)(y-2) in²

∴ A =(x-6)(y-2)

Putting the value of y

$\Rightarrow A =(x-6)(\frac{75}{x}-2)$

$\Rightarrow A = 87-\frac{450}{x}-2x$

Differentiating with respect to x

$A '= \frac{450}{x^2}-2$

Again differentiating with respect to x

$A''=-\frac{900}{x^3}$

To find the minimum area of printing space, we set A' = 0

$\therefore \frac{450}{x^2}-2=0$

$\Rightarrow 450 =2x^2$

$\Rightarrow x^2=225$

$\Rightarrow x=\pm 15$

Now putting x=±15 in A''

$A''|_{x=15}=-\frac{900}{15^3}$

$A''|_{x=-15}=-\frac{900}{(-15)^3}=\frac{900}{(15)^3}>0$

Since at x=15 , A"<0 Therefore at x=15 , the area will be minimize.

From (1) we get

$y=\frac{75}{x}$

Putting the value of x

$y=\frac{75}{15}$

   =5 in

The dimensions of the rectangular poster is 15 in by 5 in.