Answer:
Cost = $0.08
Step-by-step explanation:
Volume of the Container =[tex]10m^3[/tex]
Volume of a Cuboid =lwh
Since the length of the base is twice the width: l=2w
[tex]2w*h*w=10\\2w^{2}h=10[/tex]
Total Surface Area of a Cuboid = 2(LW+LH+WH)
Material for the base costs $10 per square meter.
Cost of the Base = $10 X LW = $(10LW)
Cost of the Sides and the top =$6(LW+2(LH+WH))=$(6LW+12LH+12WH)
Total Cost=$(10LW)+$(6LW+12LH+12WH)
=16LW+12LH+12WH
From the volume,
[tex]\\2w^{2}h=10\\h=\frac{10}{2w^{2}} =5w^{2}[/tex]
Also, l=2w
Substituting H and L into 16LW+12LH+12WH
[tex]16(2W)W+12(2W)(5W^2)+12W(5W^2)\\Cost(W)=32W^2+120W^3+10W^3\\Cost(W)=32W^2+240W^3[/tex]
The Minimum Value of C(W) is at the point where the derivative=0.
[tex]C^{'}(W)=64W+720W^2\\If C^{'}(W)=0\\64W+720W^2=0\\720W^2=-64W\\720W=-64\\W=-\frac{4}{45}[/tex]
[tex]Cost(W)=32(-\frac{4}{45})^2+240(-\frac{4}{45})^3\\Cost = $0.08[/tex]
Answer:
$165.75
Step-by-step explanation:
Let's assume the width to be = x
Since the length is twice that of the base,then the lenght will be = 2x
Let height = y
with the above variables, we set an equation for volume and cost.
10 = 2x × y = volume
Volume: V = Lwh
10 = (2x)(x)(y)
10 = 2yx²
y = 5/x²
Cost: C(w) = 10(Lw) + 2[6(hw)] + 2[6(hL)])
= 10(2x²) + 2(6(yx)) + 2(6(y)(2x)
= 20x² + 2[6w(5/x²)] + 2[12w(5/x²)]
= 20w^2 + 60/w + 120/w
= 20x² + 180x^(-1)
C'(x) = 40x - 180x^(-2)
Critical numbers:
(40x³- 180)/x² = 0
40x³-180 = 0
40x³= 180
x³= 9/2
x = 1.65 m ( the width)
2x= 3.30 m (the length)
y = 1.84 m (the height)
Cost: C = 10(Lw) + 2[6(hw)] + 2[6(hL)])
= 10(3.30)(1.65) + 2[6(1.84)(1.65)] + 2[6(1.84)(3.30)])
= $165.75 cheapest cost
The cost of materials will be cheapest when the width of the
container is 1.65 meters, base length is 3.30 meters, and the height is
1.84 meters.
