The question is in the description

Question

The top and bottom margins of a poster are 2 cm and the side margins are each 3 cm. If the area of printed material on the poster is fixed at 382 square centimeters, find the dimensions of the poster with the smallest area.

Answer 1

To solve this problem let xy=382 cm². The dimensions of the poster become

x+6 and y+4 and the area is (x+6)(y+4)=A. Substitute x=382/y and the area is (382/y+6)(y+4)=A= 382+6y+1528/y +24. So A= 406+6y+1528/y and dA/dy=6-1528/y²=0 and d²A/dy² =3056/y³ which is positive indicating a minimum since the dimensions are positive. To find y solve the equation to get 6y²= 1528 which gives y= 15.958 cm and using this to find x gives 23.94cm. o the dimensions of the poster for minimum area are x+6=29.94 and y+4=19.958.

Answer 2

Let h be the height of the printed material and w be the width of the printed material. Then the constraint is:

wh =382

We're trying to minimize the Area of the poster. The width of the poster is the width of the printed material plus the two 3cm margins on the side. Thus w + 3 + 3 or w + 6. Similarly, the height of the poster is h + 4

Thus, the area of the poster is (w+6)(h+4)