+81 1.The consumers will demand 30 units when the price of a product is $53, and 47 units when the price is $33. Find the demand function (express thr price in terms of the quantity ), assuming it is linear.
p=
2. A rectangular storage container with an open top is to have a volume of 80 cubic meters. The length of its base is twice the width. Material for the base costs 80 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.
Minimum cost =
3.The half-life of Radium-226 is 1590 years. If a sample contains 200 mg, how many mg will remain after 1000 years?
Hint: Recall \( P = P_{0}e^{kt}v\) where \( P_{0}\) is the initial amount of such a substance, P is the amount at a given time t and k is the exponential rate of decay. What would it mean if \(P = .5 P_{0}\) ?
+128406 1.The consumers will demand 30 units when the price of a product is $53, and 47 units when the price is $33. Find the demand function (express thr price in terms of the quantity ), assuming it is linear.
p=
We have the points (30, 53) and (47, 33)
We need to find the slope....so we have
[ 33 - 53] / [ 47 - 30 ] = [ -20 ] / [ 17] = -20/17
So we have
p = (-20/17)(d - 30) + 53
p = (-20/17)d + 600/17 + 53
p = (-20/17)d + 1501/17
+128406 2. A rectangular storage container with an open top is to have a volume of 80 cubic meters. The length of its base is twice the width. Material for the base costs 80 dollars per square meter. Material for the sides costs 9 dollars per square meter. Find the cost of materials for the cheapest such container.
Minimum cost =
The dimensions of the base are L * W = 2W * W = 2W^2
Ans the height is H
The volume, V, can be expressed as
80 = L * W * H
80 = 2W * W * H
80 = 2W^2 * H solving for the height, we have
40/W^2 = H
So....the surface area,SA, can be expressed as
SA = base area + area of the sides =
2W^2 + 2(40/W^2)(2W) + 2(40/W^2)(W) =
2W^2 + 240/W
So....the cost, C, can be represented as
C =
2W^2(80) + (240/W)(9) =
160W^2 + 2160/W take the derivative of this and set to 0
C ' = 320W - 2160/W^2 = 0
320W^3 - 2160 = 0
W^3 = 2160/320
W^3 = 27/4
W = 3/∛4
And the minimum cost is given by :
160(3/∛4)^2 + 2160 / (3/∛4 ) ≈ $1714.39
+128406 3.The half-life of Radium-226 is 1590 years. If a sample contains 200 mg, how many mg will remain after 1000 years?
We need to find k....[A is some original amount]
.5A = A * e^(k * 1590) divide through by A
.5 = e^(k * 1590) take the Ln of both sides
Ln (.5) = Ln e^(k * 1590) and we can write
Ln (.5) = (k * 1590)
k = Ln (.5) / 1590
So.....if the original amount is 200 mg, the amount remaining after 1000 years is
200* e^( Ln (.5) / 1590 * 1000) ≈ 129.3 mg
