w = ∛(800/120)
l = 3 * ∛(800/120)
h = 50/(3*(∛(800/120)²)
Step-by-step explanation:
First, we can say that a box has three base values -- its length, width, and height. A box also has 6 faces, with one top, one bottom, and four sides.
Each side of the box is a rectangle, and the box is a rectangular prism. Representing the width as w, length as l, and height as h, we can say that
volume = l * w * h = 50
One thing we can do is to solve for everything in terms of w so that we're only dealing with one variable. Plugging length = 3 * width into the volume equation, we get
w * 3 * w * h = 50
3w²h=50
divide both sides by 3w² to isolate h
50/(3w²) = h
In a rectangular prism, the top and bottom both have sides of lengths l and w, so the surface area/material used for those areas are each equal to w * l = w * 3 * w = 3w² . There is one top and one bottom, so we add 3w² to itself to get 6w² as the total material needed for the top and bottom.
Similarly, for the four remaining sides of the prism, two sides each have sides with lengths h and l and h and w, making the material used for the four sides equal to
2 * (h*l) + 2 * (h*w)
= 2*(50/3w² * 3*w) + 2 * (50/3w² * w)
= 2*(50/w) + 2*(50/3w)
= 100/w + 100/3w
The material used for the top and bottom costs 10 dollars per square foot, so we can multiply the surface area by the cost for material to get
6w² * 10 = 60w² as the cost for the top and bottom combined. Similarly, with sides costing $6 per square foot, we can write the total cost of the sides as
6(100/w + 100/3w) = 600/w + 600/3w = 600/w + 200/w = 800/w
Adding these up, the total cost of the box is
60w² + 800/w. We have made sure to include the facts that length = 3 * width by plugging that in as well as l*w*h=50 by plugging 50/3w² in for h. Therefore, all that's left is to minimize
60w² + 800/w
To minimize this function, we can take the derivative of it, solve for when it is equal to 0, and compare that with the minimum and maximum values of w to figure out which value for w minimizes the function.
To differentiate the function, we can separate it into two derivatives and add them up. The two parts will be 60w² abd 800/w
First, to differentiate 60w² , we can use the Power Rule to make it 2(60)(w)⁽²⁻¹⁾ = 2(60)(w)= 120w
Next, for 800/w , we can use the quotient rule to find its derivative as
800 * (1/w) d/dw = 800 * ( (d/dw (1) * w - d/dw(w) * 1) / w²) = 800 * (-1/w²) = -800/w²
We can add these up to get
120w - 800/w². Solving for 0, we have
120w-800/w² = 0
Multiplying both sides by w² to remove a denominator, we have
120w³ - 800 = 0
add 800 to both sides to isolate the w³ and its coefficient
120w³ = 800
divide both sides by 120 to isolate the w³
w³ = 800/120
cube root both sides to solve for w
w = ∛(800/120)
Comparing that with the minimum and maximum of w, the minimum is w= something very close to 0 but not 0 (because 800/w cannot have a denominator of 0) with the cost being
60w² + 800/w = 60(very close to 0)² + 800/(very close to 0) = approximately 0 + approximately ∞ = approximately ∞
At the maximum of w, which is ∞, we have
60w² + 800/w = 60 * ∞ + 800/∞ = ∞ + 0 = ∞
Therefore, this confirms the minimum at w= ∛(800/120). l = 3 * w = 3* ∛(800/120) and h=50/3w² = 50/(3*(∛(800/120)²)
