MATHEMATICS COLLEGE

jewelry box with a square base is to be built with silver plated sides, nickel plated bottom and top, and a volume of 44 cm3. If nickel plating costs $1 per cm2 and silver plating costs $2 per cm2, find the dimensions of the box to minimize the cost of the materials. (Round your answers to two decimal places.) The box which minimizes the cost of materials has a square base of side length cm and a height of cm.

Answers

Answer 1

Answer:

The box which minimizes the cost of materials has a square base of side length 4.45cm and a height of 2.22 cm.

Step-by-step explanation:

Volume of the jewellery box=44cm³

The box has a square base and is to be built with silver plated sides and nickel plated top and base.

Therefore: Volume = Square Base Area X Height = l²h

l²h=44

h=44/l²

Total Surface Area of a Cuboid =2(lb+lh+bh)

Since we have a square base

Total Surface Area =2(l²+lh+lh)

The Total Surface Area of the box =2l²+4lh

Nickel plating costs $1 per cm³

Silver Plating costs $2 per cm³

Since the sides are to be silver plated and the top and bottom nickel plated:

Therefore, Cost of the Material for the jewellery box =1(2l²)+2(4lh)

Cost, C(l,h)=$(2l²+8lh)

Recall earlier that we derived:

h=44/l²

Substituting into the formula for the Total Cost

Cost, C(l)=2l²+8l(44/l²)

C=2l²+352/l

C=(2l³+352)/l

The minimum costs for the material occurs at the point where the derivative equals zero.

C'=(4l³-352)/l²

4l³-352=0

4l³=352

Divide both sides by 4

l³=88

l=4.45cm

Recall:

h=44/l²=44/4.45²=2.22cm

The box which minimizes the cost of materials has a square base of side length 4.45cm and a height of 2.22 cm.

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The drama club is participating in a bake sale to help them raise money for their next show. They want to raise $475. They have to pay $25 to have a table at the bake sale. How many cakes will they need to sell if they charge $9 per cake?Which equation could be used to solve this problem?

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Answers

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Answers

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A rumor spreads through a small town. Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor. a. Write the differential equation satisfied by y in terms of proportionality k.
b. Find k (in units of day−1, assuming that 10% of the population knows the rumor at time t=0 and 40% knows it at time t=2 days.
c. Using the assumptions in part (b), determine when 75% of the population will know the rumor.
d. Plot the direction field for the differential equation and draw the curve that fits the solution y(0)=0.1 and y(0)=0.5.

Answers

Answer:

The answer is shown below

Step-by-step explanation:

Let y(t) be the fraction of the population that has heard the rumor at time t and assume that the rate at which the rumor spreads is proportional to the product of the fraction y of the population that has heard the rumor and the fraction 1−y that has not yet heard the rumor.

$(dy)/(dt)\ \alpha\ y(1-y)$

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where k is the constant of proportionality, dy/dt = rate at which the rumor spreads

$(dy)/(dt)=ky(1-y)\n(dy)/(y(1-y))=kdt\n\int\limits {(dy)/(y(1-y))} \, =\int\limit {kdt}\n\int\limits {(dy)/(y)} +\int\limits {(dy)/(1-y)} =\int\limit {kdt}\n\nln(y)-ln(1-y)=kt+c\nln((y)/(1-y)) =kt+c\ntaking \ exponential \ of\ both \ sides\n(y)/(1-y) =e^(kt+c)\n(y)/(1-y) =e^(kt)e^c\nlet\ A=e^c\n(y)/(1-y) =Ae^(kt)\ny=(1-y)Ae^(kt)\ny=(Ae^(kt))/(1+Ae^(kt)) \nat \ t=0,y=10\%\n0.1=(Ae^(k*0))/(1+Ae^(k*0)) \n0.1=(A)/(1+A) \nA=(1)/(9) \n$

$y=((1)/(9) e^(kt))/(1+(1)/(9) e^(kt))\ny=(1)/(1+9e^(-kt))$

At t = 2, y = 40% = 0.4

c) At y = 75% = 0.75

$y=(1)/(1+9e^(-0.8959t))\n0.75=(1)/(1+9e^(-0.8959t))\nt=3.68\ days$

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