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AlexFokin [52]
3 years ago
7

If 1900 square centimeters of material are available to make a box with a square base and an open top, find the largest possible

volume of the box.
Mathematics
1 answer:
Evgen [1.6K]3 years ago
3 0

Answer:

Volume = 7969 cubic centimeter

Step-by-step explanation:

Let the length of each side of the base of the box  be A and the height of the box be H.

Area of material required to make the box  is equal to  is A^2 + 4*A*H.

A^2 + 4*A*H = 1900

 Rearranging the above equation, we get -  

`H = \frac{(1900 - A^2)}{(4*A)}

Volume of box is equal to product of base area of box and the height of the box -  

V = A*A* H

Substituting the given area we get -

\frac{A^2*(1900 - A^2)}{4A} = \frac{(1900*A - A^3)}{4}

For maximum volume

\frac{dV}{dA} =0

\frac{ 1900}{4} - \frac{3*A^2}{4} = 0

A^2 = \frac{1900}{3}

Volume of the box

= \frac{\frac{1900}{3}*(1900 - \frac{1900}{3}) }{4 * \sqrt{\frac{1900}{3} } }

= 7969 cubic centimeter

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Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

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inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

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Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
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