Area of a box.

[tonytouch311]

An open box with a rectangular base is constructed from a rectangular pice of cardboard 16 inches wide and 21 inches long by cutting congruent squares from each corner. Find the size of a CORNER square that will produce an OPENED-TOP box with the largest possible volume.

v=wlh
w=16-2x
l=21-2x
h=??? is it x?

if so, my equation would be x(16-2x)(21-2x)?

after that...im stuck, should i worry about distributing and foil? than find the derivitive? twhat exactly does the derivitive tell me again, the minimal amount of carboard needed or the amount of the small square? than i would set that derivitive equal to 0 and if i get an answer should i get the sq root of that to tell me the true size of the square?

 

[Deleted member 4993]

An open box with a rectangular base is constructed from a rectangular pice of cardboard 16 inches wide and 21 inches long by cutting congruent squares from each corner. Find the size of a CORNER square that will produce an OPENED-TOP box with the largest possible volume.

v=wlh
w=16-2x
l=21-2x
h=??? is it x? .......YES

if so, my equation for Volume would be V(x) = x(16-2x)(21-2x)?

after that...im stuck, should i worry about distributing and foil? than find the derivitive? twhat exactly does the derivitive tell me again, the minimal amount of carboard needed or the amount of the small square? than i would set that derivitive equal to 0 and if i get an answer should i get the sq root of that to tell me the true size of the square?

What condition/s would you have to satisfy at the local extrema of a function?

 

[tonytouch311]

The first derivitive would give me that?

 

[Deleted member 4993]

The first derivative would give me that?

I suppose you meant that the condition is that:

$\displaystyle \frac{df(x)}{dx} \ = \ 0$

Then

You are correct.