Calc word problem..ouch?

Question

[A rectangular storage container with an open top is to have a volume of 10 m3. The length of its base is twice the width. Material for the base costs $10 per sq meter. Material for the sides costs $6 per sq meter. find the cost of materials for the cheapest such container. Note: you need to minimize the total cost, NOT the total amount of material used]

ok..i am a mess.

Answer 1

COST = SUM (area of piece of box)*(cost of piece/m^2)
We want to minimize cost.
We want to have cost as a function of one dimension, ie: width. We know length=2*width and that Volume/(2*width)*(width)= height.

Let w=width.
COST=(2*w^2)(10) + 2[(2*w)(V/(2w^2)]*6
+2(w*(V/2w^2)*6 [bottom+2lh+2wh]
COST=20w^2+ 36 V/2w
d(COST)/dw= 40w - 180/w^2
At minimum, derivitive =0 , so w^3= 180/40 or
w=4.5^(1/3) [about 1.7 m], length=3.4 m, and ht= 1.8 m (all dimensions appx)

Answer 2

L = 2W
H = 10 / LW = 10/ (2 W^2) = 5 / W^2

Cost of material = ($10)(LW) + 2($6)(HW) + 2($6)(HL)
= 10(2W^2) + 12(5/W) + 12(10/W)

Simplify the RHS, set dC/dW = 0 and solve for W. (Make sure that C'' is > 0, to verity that this is a local min).