Question on optimization and limits

[semc]

You are designing a rectangular poster to contain 50 cm² of printing with margins of
4 cm each at the top and bottom and 2 cm at each side. What overall dimensions will
minimize the amount of paper used?

What i did was let the length and breath of the whole poster to be x and y so the area would be 50=(x-4)*(y-8) and perimeter=2(x-4)+2(y-8). Equate the area into the perimeter and differentiate Perimeter wrt y. However i got x=y which means its the maximum area?

Limit as x tends to 0 $$\frac{e^x + e^-^x -2}{1-cos2x}$$
Applied Hopital rule once and got 0/2 however answer is 1/2. Am i correct?

 

[Mark44]

You are designing a rectangular poster to contain 50 cm² of printing with margins of
4 cm each at the top and bottom and 2 cm at each side. What overall dimensions will
minimize the amount of paper used?

What i did was let the length and breath of the whole poster to be x and y so the area would be 50=(x-4)*(y-8) and perimeter=2(x-4)+2(y-8). Equate the area into the perimeter and differentiate Perimeter wrt y. However i got x=y which means its the maximum area?

What does it mean to "equate the area into the perimeter"?

I would approach this in a different way by letting w and h represent the width and height, respectively of the printed area. From these definitiions you get wh = 50.

Now what you want to do is to minimize the area (not perimeter) of the overall piece of poster paper, so you want to minimize A = (w + 4)(h + 8). Use the other relationship to rewrite A as a function of only one variable, and then do your calculus magic.

The stuff below seems to be unrelated to this problem.

Limit as x tends to 0 $$\frac{e^x + e^-^x -2}{1-cos2x}$$
Applied Hopital rule once and got 0/2 however answer is 1/2. Am i correct?

 

[vishal007win]

for the 2nd problm... apply it once again...it will give the correct answer
after applyin it once it still gives 0/0 form...

 

[semc]

Got it guys thanks