An open rectangular box having a volume of 256 is to be constructed from a tin sheet.
Find the dimensions of such a box if the amount of material used in its construction is to be minimal.
Hint: Let the dimensions of the box be x by y by z . Then, xyz=256 and the amount of material used is given by . Show that f(x,y)= xy + 512/x + 512/y
minimize f(x,y)
What is caluculous
are open cardboard has a base of 108cm3 diameter what an the area calculate And give answer .
To find the dimensions of the box that minimize the amount of material used, we need to minimize the function f(x,y) = xy + 512/x + 512/y. We can do this by finding the partial derivatives of f(x,y) with respect to x and y, setting them equal to zero, and solving the resulting equations.
First, let's find the partial derivative of f(x,y) with respect to x:
∂f/∂x = y - 512/x^2
Next, let's find the partial derivative of f(x,y) with respect to y:
∂f/∂y = x - 512/y^2
Setting both partial derivatives equal to zero:
y - 512/x^2 = 0 (Equation 1)
x - 512/y^2 = 0 (Equation 2)
From Equation 1, we can rearrange to solve for y:
y = 512/x^2
Substituting this expression for y into Equation 2:
x - 512/(512/x^2)^2 = 0
Simplifying this equation:
x - 512/(262,144/x^4) = 0
x - x^4/512 = 0
Multiplying both sides of the equation by 512:
512x - x^4 = 0
Rearranging:
x^4 - 512x = 0
Factoring out an x:
x(x^3 - 512) = 0
Setting each factor equal to zero:
x = 0 (Extraneous solution)
x^3 - 512 = 0
Solving for x:
x^3 = 512
Taking the cube root of both sides:
x = 8
Now that we have the value of x, we can substitute it into Equation 1 to solve for y:
y = 512/x^2
y = 512/8^2
y = 8
Finally, we can substitute the values of x and y into the equation xyz = 256 to solve for z:
8 * 8 * z = 256
64z = 256
z = 4
Therefore, the dimensions of the box that minimize the amount of material used are x = 8, y = 8, and z = 4.
we know xyz = 256
so z = 256/(xy)
Surface area = SA
= 2xy + 2xz + 2yz
= 2xy = 2x(256/(xy)) + 2y(256/(xy))
= 2xy + 512/y + 512/x
hope that helps.