Find dimensions- open top box, square base, must hold 32 cubic inches, built w/minimum materials?
Let a = length of side on base of box h = height box V = volume box S = surface area box Given V = 32 Find a and h to minimize S. V = a²h h = V/a² S = a² + 4ah = a² + 4a(V/a²) = a² + 4V/a To find the critical points take the derivative of S and set it equal to zero. dS/da = 2a - 4V/a² = 0 2a = 4V/a² a = 2V/a² a³ = 2V a = (2V)^(1/3) Take the second derivative to determine the nature of the critical point. d²S/da² = 2 + 8V/a³ > 0 for all a So it is a relative minimum which is what we want. Plug into h. h = V/a² = (2V/a²)/2 = a/2 Now plug into S. S = a² + 4ah = a² + 4a(a/2) = a² + 2a² = 3a² S = 3[(2V)^(1/3)]² = 3(2V)^(2/3) S = 3(2*32)^(2/3) = 3*64^(2/3) = 3*16 = 48 sq in
Finding lengths and minimum surface area of an open rectangular box...?
Open box, so there is no top SA = 2*l*h + 2*w*h + l*w V = l*w*h = 32cm^3 h = 32/(w*l) SA = 2*l*32/(w*l) + 2*w*32/(w*l) + l*w SA = 64/w + 64/l + l*w Find minimum SA: take partial derivatives to get critical point(s) SAw = -64/w^2 + l SAl = -64/l^2 + w Both the partials have to be 0, so... 0 = -64/w^2 + l and 0 = -64/l^2 + w 64/w^2 = l 0 = -64/(64/w^2)^2 + w (plug into second equation) 0 = -w^4/64 + w 0 = w(1-w^3/64) 1 = w^3/64 or 0 = w (impossible answer) 64 = w^3 4 = w Plug w back into 64/w^2 = l 64/4^2 = l 4 = l Plug w and l back into h = 32/(w*l) h = 32/(4*4) h = 2 The Surface Area:2*l*h + 2*w*h + l*w SA = 2*4*2 + 2*4*2 + 4*4 SA = 48 So the answer is length = 4cm, width = 4cm, and height = 2cm, with Surface Area of 48cm^2
What is the surface area of a cuboid open at one end?
If the cuboid has dimensions [math]a \times b \times c[/math]then it has a total exterior surface area [math]E=2ab+2ac+2bc[/math]and a total interior surface area [math]I=2ab+2ac+2bc[/math]If the cuboid is open at one end, then we can say without loss of generality that it loses the inside and outside of one [math]ab[/math] side.So the total surface area that remains is [math]\quad S=4ab+4ac+4bc-2ab=2ab+4ac+4bc[/math]
An open top box is to be made with dimensions 3cm, 4cm, and 5cm. What is the sum of the three different surface areas of these cuboids?
An open top box with dimensions of 3 cm x 4 cm x 5 cm will have a total of(3x4)+(4x5)+(5x3) = 12+20+15 = 47 sq cm where three sides meet at one of the four vertices at the bottom.
Determine the dimensions of a squared-based, open-topped prism with a volume of 24cm^3 and a minimum surface?
a = side of square b = height a^2+4ab = surface a^2*b = volume = 24 I used a lagrangian restricted optimization method L: a^2+4ab+l(a^2b-24) FOCs are 2a + 4b +2lab = 0 4a + la^2 = 0 a^2b - 24 = 0 Initial condition us emiddle to get 4a = -la^2 --->> a=-4/l sub into top to get b = 2/l + 2 sub all into bottom to get l^3=32/22 take cubic root to get l sub into any other equation combined with a= and b= to get a = 3.53035 and b=1.9256445
How is the surface area of an open cylinder calculated?
Surface area of an open cylinder implies that the ends of the cylinder are not part of the cylinder. Only the side wall. In this case, it’s pretty simple. All you need is the radius of the cylinder to compute the circumference and the height of the cylinder.The surface area of the open cylinder is then the circumference times the height:SA = 2*pi*r * hClick on any of the links above to go to the free on-line calculator to compute the surface area of an open cylinder. If you want the total surface area of the cylinder, you have to add the two circles on the ends of the cylinder:SA = 2*pi*r * h + 2*pi*r^2
What is the surface area of an open cuboid box having each edge as 5cm?
There are several methods of doing this question .But the easiest one is given below:-First find Total Surface Area of whole cuboid which is ,Given,Length (l) = 5cmBreadth (b) = 5cmHeight (h) = 5cmTSA of cuboid = 2(l*b+b*h+h*l)= 2(5*5+5*5+5*5)= 2(25+25+25)= 2(75)= 150 sq cmNow , we know that box is open from top so we have to subtract Area of rectangle from TSA .Area of rectangle = l*b= 5*5= 25 sq cmNow,The surface area of open cuboidal box= TSA of cuboidal box - Area of rectangle= 150 - 25= 125 sq cmHence,Answer - 125 sq cm
How do you find the surface area of a rectangular box?
Hi!By surface area, I presume that the question refers to the total surface area; However, we would calculate the lateral surface (Wall) Area too.Now, let us look at the various surfaces we have; we have the front and back faces (two rectangles with sides L and H => Area of LxH each OR 2LH for both), right and the left faces (two rectangles with sides B and H => Area of BxH each OR 2BH for both),top and bottom faces (two rectangles with Sides L and B => Area of LxB each OR 2LB for both).Now, total surface area = 2LB + 2BH + 2LH = 2 (LB + BH +LH) sq.units.Talking about the wall area, we do not need the top and the bottom faces and hence, remove the 2xLxB part from the above area. The final wall area will be 2(LH + BH)= 2 x (L+B) x H sq.units.Hope this helps! :)
The total surface area of a cubical box is 96 square inches. What is the edge of the box?
The total surface area of an object is the sum of the surface area of each face, so for a cube with six equal faces this is 6A. Each face is a square of side length L, so total area is 6L^2=96. This gives L=4 inches.