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Rina8888 [55]
3 years ago
6

A box with an open top has vertical sides, a square bottom, and a volume of 108 cubic meters. if the box has the least possible

surface area, find its dimensions. (in your answer leave a space between the number and the unit.)
Mathematics
1 answer:
Colt1911 [192]3 years ago
7 0
Let x be the height  of the  box and y be the length of one side of the base then:-

V = xy^2 = 108

x = 108/y^2

Surface area = y^2 + 4xy 
S = y^2 + 4y* 108/y^2
S = y^2 + 432/y
Finding the derivative:-
dS/dy  = 2y - 432/y^2  = 0
2y^3 = 432
y^3 = 216
y = 6 

Check if this gives a minimum value:-
second derivative = 2 + 864/y^3  which is positive so  minimum.

V = xy^2 = 108
36y = 108
y = 3

Answer :-  dimensions of the box is 3*6*6 metres


 
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Pratap puri rowed 10 miles down a river in 2 ​hours, but the return trip took him 2 and one half hours. Find the rate pratap can
N76 [4]

Answer:

<em>Rate of Pratap in still water is 4.5 miles/hour and rate of current is 0.5 miles/hour.</em>

Step-by-step explanation:

Pratap Puri rowed 10 miles down a river in 2 ​hours, but the return trip took him 2.5 hours.

We know that, Speed = \frac{Distance}{Time}

So, the <u>speed of Pratap with the current</u> will be:  (\frac{10}{2})miles/hour = 5 miles/hour

and the <u>speed of Pratap against the current</u> will be:  (\frac{10}{2.5})miles/hour = 4 miles/hour.

Suppose, the rate of Pratap in still water is x and the rate of current is y.

So, the equations will be........

x+y= 5 .............................. (1)\\ \\ x-y=4 .............................. (2)

Adding equation (1) and (2) , we will get......

2x=9\\ \\ x=\frac{9}{2}= 4.5

Now, plugging this x=4.5 into equation (1), we will get.....

4.5+y=5\\ \\ y=5-4.5 =0.5

Thus, Pratap can row at 4.5 miles per hour in still water and the rate of the current is 0.5 miles/hour.

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OverLord2011 [107]
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so 1,852 is
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Answer:

C

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