Chocolate Box Company is going to make open-topped boxes out of 5 × 17-inch rectangles of cardboard by cutting squares out of the corners and folding up the sides. What is the largest volume box it can make this way? (Round your answer to the nearest tenth.)

Thank You for the help, I really appreciate it!

Damon, thank you! But i tried entering it in but it said it was wrong.

Also it wants the answer in inches cubed (in^3)

Hey, use that h, calculate L and w

find L W h

Ahhh okay my bad. Thank you!!!

I noticed an error in Damon's solution

4th line : V = (17-2h)(5-2h)h = (17-h)(5h-2h^2)
should be : V = (17-2h)(5-2h)h = (17-2h)(5h-2h^2)

I had h = 1.1445

Well, well, well, Chocolate Box Company, looks like you're up against a geometry challenge. Don't worry, I'm here to make things a little sweeter for you.

So, let's break it down. You have a 5 × 17-inch rectangle of cardboard, and you want to make open-topped boxes by cutting squares out of the corners and folding up the sides. The question is, what's the largest volume box you can create?

To solve this, we need to find the dimensions that will maximize the volume. Let's call the length of the side of the square we cut out "x". So, if we cut out squares with side length "x", the resulting box will have dimensions: length = 5 - 2x and width = 17 - 2x.

The volume of the box is given by V = length × width × height. Since the box is open-topped, we don't need to worry about the height; it will be the same as the side length we cut out, which is x.

Now, we have V = (5 - 2x) × (17 - 2x) × x. Are you following me so far?

To find the largest volume, we need to find the maximum of this function. You can do this by using calculus, but I'll let you in on a little secret: the maximum volume occurs when the derivative of this function equals zero.

Alright, let's get our clown calculators out and solve this. Differentiating the volume function with respect to x, we get:

dV/dx = -4x^3 + 44x^2 - 85x + 85.

Now, we set this derivative equal to zero and solve for x to find the critical points. Pretty exciting stuff, right?

Once we find those critical points, we can plug them back into the volume function to find the maximum volume. And voila! That'll be your answer, my chocolaty friend.

Just remember to round your answer to the nearest tenth. And if you need any more help, don't hesitate to ask. Good luck, Chocolate Box Company! May your boxes be filled with joy and chocolatey delights!

To find the largest volume box that can be made using the given method, we need to employ the process of optimization by finding the maximum value of the volume function.

Let's assume that when squares are cut out from the corners, each side of the square has a length of "x" inches. Thus, the dimensions of the box will be (17-2x) inches by (5-2x) inches.

The volume of the box can be calculated by multiplying these dimensions with the height of the box, which is "x" inches.

Volume = (17-2x)(5-2x)(x)

To find the largest volume, we need to find the critical points of the volume function. To do this, we will take the derivative with respect to "x" and set it to zero:

Volume' = 0

Differentiating the volume function yields:

d/dx [(17-2x)(5-2x)(x)]

Using the product rule, this differentiation becomes:

(5-2x)(x)*(-2) + (17-2x)(x)*(-2) + (17-2x)(5-2x)*(1) = 0

Simplifying this equation, we get:

-4x^2 + 10x - 2x^2 + 34x -4x^2 + 36x -20x + 40x^2 = 0

Combining like terms:

20x^2 + 36x - 20 = 0

Now, we can solve this quadratic equation for "x" using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this equation, a = 20, b = 36, and c = -20. Plugging these values into the quadratic formula:

x = (-(36) ± √((36)^2 - 4(20)(-20))) / (2(20))

Simplifying further:

x = (-36 ± √(1296 + 1600)) / 40

x = (-36 ± √2896) / 40

x = (-36 ± 53.789) / 40

x ≈ (-36 + 53.789) / 40 or x ≈ (-36 - 53.789) / 40

x ≈ 0.443 or x ≈ -2.175

Since the length cannot be negative, we discard the negative solution.

Therefore, the value of "x" that maximizes the volume is approximately 0.443 inches.

Using this value of "x", we can calculate the dimensions of the box:

Length = 17 - 2x ≈ 17 - 2(0.443) ≈ 16.113 inches
Width = 5 - 2x ≈ 5 - 2(0.443) ≈ 4.114 inches
Height = x ≈ 0.443 inches

Finally, we calculate the maximum volume by substituting these values into the volume function:

Volume ≈ (16.113)(4.114)(0.443) ≈ 29.463 cubic inches

So, the largest volume box that the Chocolate Box Company can make using the given method is approximately 29.463 cubic inches.

L = 17 - 2 h

W = 5 - 2 h

V = L W h
V = (17-2h)(5-2h)h = (17-h)(5h-2h^2)

dV/dh = (17-h)(5 - 4h) +(5h-2h^2)(-1)

= 85 - 73 h +4 h^2 -5h + 2 h^2

= 85 - 78 h + 6 h^2
where is that derivative zero?
h = [ 78 +/- sqrt (6084-2040) ]/12

h = [ 78 +/- 63.6 ] / 12

h = 11.7 impossible, plank not wide enough
or
h = 1.2 inches