Question

which is correct?

Answer 1

?

Answer 2

The length of a cedar chest is twice its width. The cost/dm^2 of the lid is four times the cost/dm^2 of the rest of the cedar chest. If the volume of the cedar chest is 1440 dm^3, find the dimensions so that the cost is a minimum. Begin by drawing a diagram sketch.

Answer 3

x = one base length
2x = the other
y = height

then
2x^2 y = 1440
y = 720/x^2

Cost
C = A(2x^2 + 2xy + 4xy) + 2A(2x^2)

now, A is constant and y needs to be replaced
so we have C(x) with parameter A

get C'(x) = 0 and solve for such x.

Answer 4

hater

Answer 5

just playin wht you need help with

Answer 6

lol

Answer 7

hold on let me take a look

Answer 8

ok

Answer 9

which picture is correct?

Answer 10

1