Question

a 39-inch by 104-inch piece of cardboard is used to make an open-top container by removing a square from each corner of the cardboard and folding up the flaps on each side. what size square should be cut from each corner to get a container with the maximum volume? enter the area of the square and do not include any units in your answer.

Answer 1

a 39-inch by 104-inch piece of cardboard is used to make an open-top container by removing a square from each corner of the cardboard and folding up the flaps on each side. what size square should be cut from each corner to get a container with the maximum volume? enter the area of the square and do not include any units in your answer.

Answer 2

According to the solving the area of the square is 68.0625.

What's a square's area?

As is common knowledge, a square is a four-sided, two-dimensional figure. It is also referred to as a quadrilateral. The total quantity of unit squares forming a square is referred to as the square's area. In other words, it is described as the area that the square takes up.

According to the given data:

The box formed after cutting the square from each corner will have the dimensions as,

length = 104 - 2x, width = 39 - 2x, height = x.

∴ volume of the box = length × width × height

∴ v = (104 - 2x)(39 - 2x)(x) -----(i)

∴ v = (104 - 2x) (39x - 2$x^{2}$)

∴ v = 104(39x - 2$x^{2}$) -2x(39x - 2$x^{2}$)

∴ v = 4056x - 208$x^{2}$ - 78$x^{2}$ + 4$x^{3}$

∴ v = 4$x^{3}$ - 286$x^{2}$  + 4056x

let f(x) =  4$x^{3}$ - 286$x^{2}$  + 4056x  -----(ii)

To, find x for which f(x) is maximum,

⇒we should apply second derivative test ,

According to this test, first we should find critical points at which f'(x) = 0.

then if f''(x) < 0 for that critical point then f(x) is maximum at that critical point.

∴ let us consider, f'(x) = 0.

now, f(x) =  4$x^{3}$ - 286$x^{2}$  + 4056x

⇒ f'(x) = 12$x^{2}$ - 572x + 4056.  -----(iii)

⇒ f'(x) = 4(3$x^{2}$ - 143x + 1014)

⇒ f'(x) = 0.

⇒  f'(x) = 4(3$x^{2}$ - 143x + 1014) = 0

⇒ x = (-b ± $\sqrt{b^{2} - 4ac }$)/2a    ; where a = 3, b = -143, c = 1014.

∴ x = (-(-143) ± $\sqrt{(-143)^{2} -4(3)(1014)}$)/2×3

∴ x = (143 ± $\sqrt{8281}$)/6.

∴ x = $\frac{143 + 91}{6}$  , x = $\frac{143 - 91}{6}$

⇒ x = 39, 8.25

the square of length 8.25 inch should be cut from each side to det contained with the maximum volume.

Area of the square is = $x^{2}$ = $8.25^{2}$

∴ Area = 68.0625

know more about area of square visit this link:

https://brainly.com/question/1658516

#SPJ4