The top and bottom margins of a poster are each 6 cm and the side margins are each 4 cm. if the area of printed material on the poster is fixed at 384 cm , find the dimensions of the poster with the smallest area.

  24 cm wide by 36 cm high

Step-by-step explanation:

The poster with the smallest area will have an aspect ratio that makes the margin dimensions the same percentage of overall dimension in each direction.

Since the ratio of margin widths is 6:4 = 3:2, the poster and printed area will have an aspect ratio of 3:2. That is, the width is ...

  width of printed area = √(2/3·384 cm²) = 16 cm

Then the width of the poster is ...

  width = left margin + printed width + right margin = 4cm + 16 cm + 4 cm

  width = 24 cm

The height is 3/2 times that, or 36 cm.

The smallest poster with the required dimensions is 24 cm wide by 36 cm tall.

If you need to see the calculus problem, consider the printed area width to be x. Then the printed height is 384/x and the overall dimensions are ...

  (x + 8) by (384/x + 12)

We want to minimize the area, which is the product of these dimensions:

  a = (x +8)(384/x +12) = 384 +12x +3072/x +96

  a = 12x + 3072/x +480

This is a minimum where its derivative is zero.

  a' = 12 -3072/x^2 = 0

  a' = 1 -256/x^2 = 0 . . . . . . divide by 12; true when x^2 = 256

This has solutions x=±16, of which the only useful solution is x=16.

10a^3+277a^2-132a-792

step-by-step explanation:

(2a+11)(5a-12)(foil)

=10a^2-24a+55a-132 =10a^2+31a-132

(10a^2+31a-132)(a+6)

=10a^3+60a^2+31a^2+186a^2-132a-792 =10a^3+277a^2-132a-792

i hope this is what you are looking for : )