Calculus problem... I just missed on a test !! Will someone explain??

Question

A rectangular storage container with an open top is to have a volume of 10 cubic meters. The length of its base is twice the width. Material for the base costs $12 per square meter. Material for the sides costs $10 per square meter. Find the cost of the materials for the cheapest such container.

Answer 1

Area of Base = W*L = W*2W since Length is twice the Width
so Area of Base = 2W^2

There are two pairs of sides.
One pair has area = W*H
The other pair has area = L*H = 2W*H
The total area of the sides is then twice the sum of these or 6*W*H

Now Volume = 10 = W*L*H = W*2W*H= 2*W^2*H
Solve this for H and get H = 5/W^2
Use this for the area of the sides and get 30/W

Now for the Cost:
The bottom Cost = 12*[2*W^2] = 24W^2
The sides Cost = 10*[30/W[ = 300/W
Total Cost is the sum of these = 24W^2 + 300/W

You want to find what value of W gives a mninimum of Cost so you must differentiate the Cost and set it equal to 0.

d/dW[ 24W^2 + 300/W] = 48W - 300/W^2 = 0
And solve for W: W^3 = 300/48 = 150/24 = 75/12 =25/4
so W = 1.842

Substitute back into the formula for the Cost and get: 244.30 (to the nearest cent)

Answer 2

Let
x = width base
2x = length base
h = height
V = volume
S = surface area
C = cost

V = x*2x*h = 2x²h = 10
S = x*2x + 2xh + 2*2x*h = 2x² + 2xh + 4xh = 2x² + 6xh
C = 12*2x² + 10*6xh = 24x² + 60xh

V = 2x²h
h = V/(2x²)

Substitute into the equation for C.

C = 24x² + 60xh = 24x² + 60x[V/(2x²)]
C = 24x² + 30V/x

Take the derivative and set equal to zero to find the critical values.

dC/dx = 48x - 30V/x² = 0
48x = 30V/x²
48x³ = 30V
x³ = 30V/48 = 5V/8
x = (5V/8)^(1/3) = (5*10/8)^(1/3) = [50^(1/3)] / 2

Calculate the second derivative to determine the nature of the critical point.

d²C/dx² = 48 + 60V/x³ > 0 Since all terms are positive.
This indicates a minimum value which is what we want.

Plug in for x to calculate the lowest cost.

C = 24x² + 30V/x
C = 24{[50^(1/3)] / 2}² + 30*10 / {[50^(1/3)] / 2}
C = 6 [50^(2/3)] + 600 / [50^(1/3)]
C = {6 / [50^(1/3)]} (50 + 100)
C = 900 / [50^(1/3)]
C = $244.30

Answer 3

Let a= width of base
then length of base = 2 x width = 2a
Base area = 2a^2
Let height of container = h
then volume = 2a^2 h = 10, ```` so h = 1/(5a^2)
Cost of base = Base area x cost = (2a^2) x $12 = $24a^2

side area = base perimeter x height = 3a x 2 x h
= 6a x 1/(5a^2) = 6/(5a)
Cost of sides = Side area x cost = 6/(5a) x 10 = $12/a

Total Cost C(a) = 24 a^2 + 12/a
To minimize C(a), make its derivative C'(a)=0
C'(a) = 48 a - 12/(a^2) = 0 when
48 a = 12/(a^2) or a^3 = 0.25
so a = 0.63 m

and cost = $C = 24 (0.63)^2 + 12/0.63 = $79.53

Answer 4

let the width be 'w' length '2w' and height 'h' substituting h=5/w^2.
total cost is 24w^2(floor) +(2*2w*h+2*w*h)*10(sides)=24w^2+6*w*h*10
that is 24*w^2+300/w
differentiating this i got 48*w-300/w^2 putting it to be zero w comes out to be 5/2
so width is 2.5 m

Answer 5

10=l*b*h=> 10=2*h*b^2
=>5=hb^2=>h=5/(b^2)
cost=surface area*cost
=12*(l*b)+10*(2*((b*h)+(2b*h)))
=12*(l*b)+10*6*b*h
cost =24b+60bh
cost=24b+60h
cost=24b+300/b^2
differentiate with respect to b and equating with 0 => 0=24+300*-2b^-3
0=24-600b^-3
24/600 = b^-3
600/24 = b^3
b=cuberoot(25)
b approx =2.924
therefore cost=70.176+35.088638579537054538037019917247
=105.2646

Answer 6

it is obviously fake. Sin(?x / ?x may be canceled to grant us Sin(0 using the unit circle (or calculator) will provide us 0. nonetheless you could substitute a quantity into the equation. Sin2? / 2? which additionally does not equivalent a million.