Question

Please answer with work

Solve the problem. 19) From a thin piece of cardboard 30 in. by 30 in., square corners are cut out so that the sides can be folded up to make a box. a) What dimensions will yield a box of maximum volume? b) What is the maximum volume? Round to the nearest tenth, if necessary.

Answer 1

19 Given cardboard size 30 in x 30 in 30-2x k Y. 2 tet squares of size “r’in are cut in the corners, a 3oin 30-24 x K 3oin 키 When folded up to make box 71x height of the box = x in width of the box = (30-22) in length of the bod=(30-2x) in -- volume of boxe length x width x height V = (30-2x) (30-21) a = (30-2x)ºx. V = (900+ 4x-120x) x → V = 4263-1204+9002 V = For maximising volume du ax =0 ď (4x?-120 x’+9001) = 0 → 147(3) ’ –120127 x +900 (1)=0 da 2 12 x - 240 X + 900 =0 ñ ñ 2 a-202 + 75=0

x=15 length we get x = 5, 15 when - width = 30-2(15) length = width = 0 (Not possible) x=5 iu valid. DIMENSIONS :- length of the box = = 30 - 2(5) = 30-10 = 20 inches width of the boxt = 30-2(5) 20 inches Height the box= 5 inches. of (6 Maximum volume 20 X 20 X 5 2000 in3 Am