Solve Volume and Surface Area for Rectangular Open-Top Box: Apps of Cubics

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The discussion focuses on solving mathematical problems related to the volume and surface area of geometric shapes, specifically a rectangular open-top box and an open cylindrical can. The surface area of the box is derived as A = x² + 80/x, where x is the side length of the base. For the cylindrical can, the height h is expressed in terms of the radius r as h = (25 - r²) / (2r), leading to a volume formula of V = πr(25 - r²). Additionally, the discussion covers the derivation of dimensions for a rectangular box with a diagonal of 6m, resulting in the volume expressed as V = 36w - w³.

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A rectangular box with height hm and square base of side length xm is open at the top. the volume of the box is 20m

4. (b) show that the surface area A m^2 of the box is given by A=x^2 + 80/x.

* I got h in terms of x for 4(a) that was easy it is h=20/x^2 and it is right from the answer. just how do i find surface area, because i know it is the addition of all the area of the sides but i got it wrong.

5.AN open cylintrical can of radius r cm and height hcm is to be made out of 25 pi(cm)^2 of sheet tin.
(a) express the height hcm in terms of r.
I got 25pi - pi(r)^2 /2pi(r) = h
the back of the book got h = 25-r^2 / 2r someone guide me thorugh on the steps on how to get the answer.

(b) express volume in terms of r.
i know volume = area * h. and to do that i need pi(r)^2 * h from previous answer. so someone show me how to do that? because i got v=25pi(r)-pi(r)^3 2 i know that is the wrong answer because the back of the book got v=(1/2)r(25-r^2) someone show me how to get it thanks.

9. A skeleton cylinder is made of two identical circular pieces of wire joined by two identical striaght pieces the legnth of the wire used is 32cm and the radius of the cylinder is r cm.

(a) express the height hcm in terms of r
* well i am giving a radium and the side lengths, but i got condused and i did 32 + 2pi(r)^2 + 2h. I am really lost but the answer is
h=2(8-pi(r))

(b) express the volume vcm^3 of the cylinder in therms of r. state restrictions

Ok i can only do that if i have h, well from if someone tells me how to get the previous answer, i just dop v = a * h and rea is just (pi)r^2 * 2? because there are 2 cylinder heads. the restrictions is easy just to see the numbers which are in the positive maxium etc.

i tried my best someone please answer how to get the h= thanks.

10. A box has a rectangular base the diagnoal of which is 6m
(a) if the width of the base is wm, express length, lm of the base in terms of w

well i know that i have to get it in the form of l=
and, i suppose 6m is the side diagnol. i got ripped and did l^2 + 36 + w^2 I'm really lost someone please help me on how they got l=sqr(36-w^2)

(b) if the height of the box is equal to the length of the base, express the volume vm^3 in terms of w. state any restrictions on w.

* if height is equal to length, then volme = area * h where area is (l*w) and length from above * width but i don't know how to solve this. someone help me?

* SORRY TO BOTHER YOU PEOPLE BUT I AM REALLY STUCK PLEASE SHOW ME WHAT TO DO BECAUSE I KNOW I DID IT WRONG THANKS!
 
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"A rectangular box with height hm and square base of side length xm is open at the top. the volume of the box is 20m

4. (b) show that the surface area A m^2 of the box is given by A=x^2 + 80/x. "

The problem says that box is "open at the top" so it has 5 sides, not 6. The base is "x by x" so its area is x2. Each of the 4 sides is "x by h" and so has area xh= x(20/x2)= 20/x. The total are is the area of the base plus 4 times the area of each side: x2+ 4(20/x)= x2+ 80/x.

"5.AN open cylintrical can of radius r cm and height hcm is to be made out of 25 pi(cm)^2 of sheet tin.
(a) express the height hcm in terms of r."

An "open cylindrical can"? I would interpret that as meaning open at both ends (simply bend the rectangular sheet into a circle and weld opposite sides) but then I don't get the answer you give from your book). If "open" means that it has a bottom but not a top then we can argue: the area of the top is [pi] r2 so that
we have 25 [pi]- [pi]r2 left for the curved side. Assuming that we can "patch" odd shaped pieces together so that we can use all of that area we must have the area of the sides: 2[pi]rh equal to that: 2[pi]rh= [pi](25- r2) so
h= (25-r2)/2r

"(b) express volume in terms of r.
i know volume = area * h. and to do that i need pi(r)^2 * h from previous answer. so someone show me how to do that? because i got v=25pi(r)-pi(r)^3 2 i know that is the wrong answer because the back of the book got v=(1/2)r(25-r^2) someone show me how to get it thanks."

Well, yes, using the answer from (a) that you KNOW was wrong will give you the wrong answer to (b)! You told us the book gave
h= (25-r2)/2r for (a) so the volume must be
V= [pi]r2h= [pi]r2(25-r2)/2r
= [pi]r(25-r2)

"9. A skeleton cylinder is made of two identical circular pieces of wire joined by two identical striaght pieces the legnth of the wire used is 32cm and the radius of the cylinder is r cm.

(a) express the height hcm in terms of r
* well i am giving a radium and the side lengths, but i got condused and i did 32 + 2pi(r)^2 + 2h. I am really lost but the answer is
h=2(8-pi(r))"

Again, you have confused units. "32" is from 32 cm and is a length. 2 pi r2 is in "cm2" and is an area. You can't add them! Here there is no area- this is made of wire and length is the only quantity we have. A circle of radius r has circumference 2 pi r so the two circles have length 4 pi r. There are also 2 lengths of wire representing the length of the "cylinder" so their length will be 2h. The total length of the wires is
4 pi r+ 2h which must equal 32 cm. 4 pi r+ 2h= 32 so
2h= 32- 4 pi r and h= 16- 2 pi r= 2(8- pi r).

"(b) express the volume vcm^3 of the cylinder in therms of r. state restrictions

Ok i can only do that if i have h, well from if someone tells me how to get the previous answer, i just dop v = a * h and rea is just (pi)r^2 * 2? because there are 2 cylinder heads. the restrictions is easy just to see the numbers which are in the positive maxium etc."

The area is pi r2, not two times that (you would count both top and bottom if you were calculating total area, not volume).

The volume is [pi] r2h= [pi]r2(16- 2 pi r).

Since all measurements must be positive, r must be greater than 0 and 16- 2 pi r must be greater than 0 which means r must also be less than 8/pi.

"10. A box has a rectangular base the diagnoal of which is 6m
(a) if the width of the base is wm, express length, lm of the base in terms of w

well i know that i have to get it in the form of l=
and, i suppose 6m is the side diagnol. i got ripped and did l^2 + 36 + w^2 I'm really lost someone please help me on how they got l=sqr(36-w^2)"

? What is a "side" diagonal"? You are told directly that the diagonal of the base is 6 meters. What you were trying to remember is the Pythagorean theorem: a2+b2= c2 for a right triangle. Since the diagonal of a rectangle is the hypotenuse of the right triangle formed by the sides, you have
l2+w2= 62= 36.
That leads to l2= sqrt(36- w2).

"(b) if the height of the box is equal to the length of the base, express the volume vm^3 in terms of w. state any restrictions on w.

* if height is equal to length, then volme = area * h where area is (l*w) and length from above * width but i don't know how to solve this. someone help me?"

Now the volume is length* width* height and since height= length
V= l2w. From (a), w= sqrt(36-w2) so
V= (36-w2)w= 36w- w3.
 

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