Polynomial Functions

[ender7x77]

K i got this application question the other day in my introduction to calculus class and i was wondering if anyone could lend me a hand.

 

Katie is building a wooden rectangular storage box. The box will have an open top and a volume of 1.5m cubed. For design purposes, Katie would like the length of its base to be triple its width. Thick wood for the base costs $8/m squared and thinner wood for the sides costs $5/m squared.

a) Express the cost of the wood as a finction of the width of the base.

b) Find all possible dimensions if Katie spends $44 for the wood.

c) what dimensions would you recommend? Why?

 

Now, i think if someone could help me with a) then i probably would be able to get the rest. I have figured out the base of the box is 3w + w, where w represents the the width size, but given that i do not know how to get the height of the box. I believe if i can determine the w variable then i could find the height by manipulating the formula for the volume of a box (V= lwh). anyways any help would be greatly appreciated..

[ender7x77]

oh ya when i said the volume was 1.5m cubed, i didn't mean to the exponent and i am sorry for the confusion...the same thing for the cost of the wood. anyways plz help i am totally lost

[woelen]

In order to solve this problem the assumption is made that the thickness of the wood can be neglected, so we simply work with width w and not with w plus a few mm for the wood's thickness.

 

First compute the height as a function of w.

 

The area of the base is 3w². The total volume must be 1.5 m³. So, the height is 1.5/(3w²) = 0.5/w².

 

Now, what is the area of the sides? There are four sides. Each side has a height of 0.5/w² and two of them have a length w, and two of them have a length 3w. So, the total area of wood for the sides is 2*3w *(0.5/w²) + 2*w*(0.5/w²), which is 3/w + 1/w = 4/w.

 

So, the cost of the box is 5*(4/w) + 8*3w². = 20/w + 24w².

 

E.g. for a box with a width of 2 meters, the cost would be 20/1 + 24*1*1 = 20+24 = 44 dollars.

 

This is the answer to question (a). Now try to obtain the answers for (b) and ©. You have seen the reasoning and now the others should be possible for you.

[ender7x77]

In order to solve this problem the assumption is made that the thickness of the wood can be neglected, so we simply work with width w and not with w plus a few mm for the wood's thickness.

 

First compute the height as a function of w.

 

The area of the base is 3w². The total volume must be 1.5 m³. So, the height is 1.5/(3w²) = 0.5/w².

 

Now, what is the area of the sides? There are four sides. Each side has a height of 0.5/w² and two of them have a length w, and two of them have a length 3w. So, the total area of wood for the sides is 2*3w *(0.5/w²) + 2*w*(0.5/w²), which is 3/w + 1/w = 4/w.

 

So, the cost of the box is 5*(4/w) + 8*3w². = 20/w + 24w².

 

E.g. for a box with a width of 2 meters, the cost would be 20/1 + 24*1*1 = 20+24 = 44 dollars.

 

This is the answer to question (a). Now try to obtain the answers for (b) and ©. You have seen the reasoning and now the others should be possible for you.

 

thanks a lot...i'm gonna get some sleep cuz i had a long day and try it tomorrow...thanks once again i was beginning to think no one was gonna respond