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hammer [34]
3 years ago
14

A sells to b an item at 15% profit. B sells the same item to c at 20% profit. If c pays $1656 for it. What is the price at which

a bought the item?
Mathematics
1 answer:
andrew11 [14]3 years ago
8 0

Answer:

<em>The price at which A bought the item is $1200.</em>

Step-by-step explanation:

Suppose, the price at which A bought the item is  x dollar.

A sells the item to B at 15% profit. That means, the selling price to B = x+(0.15*x)=1.15x dollar.

B sells the same item to C at 20% profit. That means, the selling price to C = 1.15x+(1.15x*0.20)=1.38x dollar.

Given that, C pays $1656 for the item. So, the equation will be.......

1.38x=1656\\ \\ x=\frac{1656}{1.38}=1200

So, the price at which A bought the item is $1200.

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MARSVECTORCALC6 3.4.020. My Notes A rectangular box with no top is to have a surface area of 64 m2. Find the dimensions (in m) t
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Answer:

We would have

                                    l =w =\frac{8\sqrt{3}}{3} \\h = \frac{8\sqrt{3}}{6}

where " l " is  length, " w"  is width and "h" is height.

Step-by-step explanation:

Step 1

Remember that

         Surface area for a box with no top = lw+2lh+2wh = 64

where " l " is  length, " w"  is width and "h" is height.

 Step 2.

Remember as well that

                              Volume of the box = l*w*h

Step 3

 We can now use lagrange multipliers.  Lets say,

                                    F(l,w,h) = lwh

and

                                g(l,w,h) = lw+2lh+2wh = 64

By the lagrange multipliers method we know that                            

 

                                                     \nabla F  = \lambda \nabla g

Step 4

Remember that

                          \nabla F  = (wh,lh,lw)

and

                      \nabla g = (w+2h,l+2h , 2w+2l)

So basically you will have the system of equations

                              wh = \lambda (w+2h)\\lh = \lambda (l+2h)\\lw = \lambda (2w+2l)

Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get

                                   lwh = l\lambda (w+2h)\\\\lwh = w\lambda (l+2h)\\\\lwh = h\lambda (2w+2l)

Then you would get

                      l\lambda (w+2h) = w\lambda (l+2h) =  h\lambda (2w+2l)

You can get rid of \lambda from these equations and you would get

                         lw+2lh = lw+2wh =  2wh+2lh

And from those equations you would get

                                             l = w =2h

Now remember the original equation

                                    lw+2lh+2wh = 64

If we plug in what we just got, we would have

                                 l^{2} + l^{2} + l^2   =  64 \\3l^{2} = 64 \\l = w = \frac{8\sqrt{3} }{3} \\h = \frac{8\sqrt{3} }{6}

                       

                                                 

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