Squares with sides of length x are cut out of each corner of a rectangular piece of cardboard measuring 3737 ft by 2020 ft. The resulting piece of cardboard is then folded into a box without a lid. Find the volume of the largest box that can be formed in this way. b. Suppose that in part​ (a) the original piece of cardboard is a square with sides of length s. Find the volume of the largest box that can be formed in this way. a. To find the objective​ function, express the volume V of the box in terms of x.

Respuesta :

Answer:

a) The volume of the largest box that can be formed is [tex]1,393.53 ft^3[/tex].

b) The maximum volume of the box: [tex]\frac{2s^3}{27}[/tex]

Explanation:

a) Length of the card board = l = 37 ft

Width of the card board= b = 20ft

Squares with sides of length x are cut out of each corner of a rectangular cardboard to form a box.

Now, length of the box = L = 37 -2x

Breadth of the box ,B= 20- 2x

Height of the box ,H= x

Volume of the box ,V= L × B × H

[tex]\frac{dV}{dx}=\frac{(37 -2x)(20-2x)x}{dx}[/tex]

[tex]\frac{dV}{dx}=\frac{d(740x-114x^2+4x^3)}{dx}[/tex]

[tex]\frac{dV}{dx}=740-228x+12x^2[/tex]

Putting ,[tex]\frac{dV}{dx}=0[/tex]

[tex]0=740-228x+12x^2[/tex]

x = 4.1537 , 14.846

[tex]\frac{d^2V}{(dx)^2}=-228+24x[/tex]

When , x = 4.1537 , [tex]\frac{d^2V}{(dx)^2}<0[/tex] (maxima)

The volume of the largest box that can be formed:

[tex](27 -2x)(15-2x)x = (37 -2\times 4.1537 )(20-2\times 4.1537 )\times 4.1537 [/tex]

[tex]V=1,393.53 ft^3[/tex]

The volume of the largest box that can be formed is [tex]1,393.53 ft^3[/tex].

b) Original piece of cardboard is a square with sides of length s.

Length of the card board = l = s

Squares with sides of length x are cut out of each corner of a rectangular cardboard to form a box.

Now, length of the box = L = s -2x

Height of the box ,H = x

Volume of the box ,V= L × L × H

[tex]=\frac{dV}{dx}=\frac{(s -2x)(s-2x)x}{dx}[/tex]

[tex]\frac{dV}{dx}=\frac{d((s-2x)^2x)}{dx}[/tex]

[tex]\frac{dV}{dx}=(s-2x)^2+x(2(s-2x)(-2))[/tex]

[tex]\frac{dV}{dx}=(s-2x)^2-8x(s-2x)[/tex]

[tex]\frac{dV}{dx}=O[/tex]

[tex]x=\frac{s}{6},\frac{s}{2}[/tex]

[tex]\frac{d^2V}{(dx)^2}=2(s-2x)(-2)-8(2-2x)+16x[/tex]

When , [tex]x=\frac{s}{6} , \frac{d^2V}{(dx)^2}<0[/tex] (maxima)

The maximum volume of the box:

[tex]V_{max}=(s-2x)^2x=(s-2\times \frac{s}{6})^2\times \frac{s}{6}=\frac{2s^3}{27}[/tex]