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word problem

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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 \(P = .5 P_{0}\) ?

Jun 7, 2018
edited by Sloan  Jun 7, 2018
edited by Sloan  Jun 7, 2018

#1
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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   Jun 8, 2018
#2
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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   Jun 8, 2018
#3
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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   Jun 8, 2018