A rectangular sheet of tin-plate is 2k cm by k cm. Four squares, each with sides x cm, are cut from its corners. The remainder is bent into the shape of an open rectangular container. Find the value of x which will maximize the capacity of the container.
I know the volume is x(2k-2x)(k-2x) but don't know where to go from there
keep in mind that k will be a constant, so ..
V = x(2k^2 - 4kx - 2kx + 4x^2)
= (2k^2)x - (6k)x^2 + 4x^3
dV/dx = 2k^2 - 12kx + 12x^2 = 0 for a max of V
x = (12k ± √(144k^2 - 4(12)(2k^2))/24
= k(12 ± √48)/24
= k(12 ± 4√3)/24
= k(3 ± √3)/8 , where k > 2x
Going from
2k^2 - 12kx + 12x^2 = 0
I get a divisor of 4, not 24
I read the quadratic backwards, for some strange reason!
To find the value of x that maximizes the capacity of the container, we need to find the critical points of the volume function and determine which one gives the maximum value.
Let's start by expressing the formula for the volume:
V(x) = x(2k - 2x)(k - 2x)
To find the critical points, we need to differentiate V(x) with respect to x and set the derivative equal to zero:
V'(x) = 0
Taking the derivative of V(x) using the product rule, we have:
V'(x) = (2k - 2x)(k - 2x) + x(-2)(k - 2x) + x(2)(2k - 2x)
Now let's simplify and solve for x:
0 = (2k - 2x)(k - 2x) - 2x(k - 2x) + 4x(2k - 2x)
0 = (2k - 2x)(k - 2x) - 2kx + 4x^2 + 4kx - 8x^2
0 = 2k^2 - 4kx + 2x^2 - 2kx + 8x^2
0 = 2k^2 + 6x^2 - 6kx
Now, we can simplify further:
2k^2 + 6x^2 = 6kx
k^2 + 3x^2 = 3kx
k^2/3 + x^2 - kx = 0
Now we need to solve this equation, but it doesn't seem trivial to solve directly for x. However, we can take advantage of the fact that the container's dimensions cannot be negative, so x must be positive.
To simplify the equation and make it more approachable, let's substitute a new variable:
y = x/k
We can rewrite the equation as:
(k^2/3) + (k^2)y^2 - k^2y = 0
Now we can solve for y:
k^2(1/3 + y^2 - y) = 0
This equation yields two solutions:
1) y = 0
2) 1/3 + y^2 - y = 0
For y = 0, it corresponds to the minimum value of the volume since it would result in a flat piece of tin without any container shape.
Now let's solve the equation 1/3 + y^2 - y = 0:
y^2 - y + 1/3 = 0
Using the quadratic formula, we have:
y = (-b ± √(b^2 - 4ac)) / (2a)
where a = 1, b = -1, and c = 1/3
Solving for y, we get:
y = (1 ± √(1 - 4(1)(1/3))) / 2
y = (1 ± √(1 - 4/3)) / 2
y = (1 ± √(3/4)) / 2
y = (1 ± √(3)/2) / 2
Since y = x/k, we can write:
x/k = (1 ± √(3)/2) / 2
Simplifying further, we find:
x = k(1 ± √(3)/2) / 2
So the two critical points are:
1) x = k(1 + √(3)/2) / 2
2) x = k(1 - √(3)/2) / 2
To determine which value of x maximizes the volume, we need to evaluate the volume function V(x) at these two critical points:
V(1) = 1(2k - 2)(k - 2) = 2k(2-k)
V(2) = 2(2k - 4)(k - 4) = 4(4k-8)(k-4) = 16-32k+16k^2-32k+64
V(3) = 3(2k - 6)(k - 6) = 6(6k-36)(k-6) = 36-216k+216k^2-216k+1296
We can now compare these value to determine which one gives the maximum volume.