Respuesta :
Answer:
[tex]\frac{10}{3} \ inches[/tex] of square should be cut out of each corner to create a box with the largest volume.
Step-by-step explanation:
Given: Dimension of cardboard= 16 x 30“.
As per the dimension given, we know Lenght is 30 inches and width is 16 inches. Also the cardboard has 4 corners which should be cut out.
Lets assume the cut out size of each corner be "x".
∴ Size of cardboard after 4 corner will be cut out is:
Length (l)= [tex]30-2x[/tex]
Width (w)= [tex]16-2x[/tex]
Height (h)= [tex]x[/tex]
Now, finding the volume of box after 4 corner been cut out.
Formula; Volume (v)= [tex]l\times w\times h[/tex]
Volume(v)= [tex](30-2x)\times (16-2x)\times x[/tex]
Using distributive property of multiplication
⇒ Volume(v)= [tex]4x^{3} -92x^{2} +480x[/tex]
Next using differentiative method to find box largest volume, we will have [tex]\frac{dv}{dx}= 0[/tex]
[tex]\frac{d (4x^{3} -92x^{2} +480x)}{dx} = \frac{dv}{dx}[/tex]
Differentiating the value
⇒[tex]\frac{dv}{dx} = 12x^{2} -184x+480[/tex]
taking out 12 as common in the equation and subtituting the value.
⇒ [tex]0= 12(x^{2} -\frac{46x}{3} +40)[/tex]
solving quadratic equation inside the parenthesis.
⇒[tex]12(x^{2} -12x-\frac{10x}{x} +40)[/tex]=0
Dividing 12 on both side
⇒[tex][x(x-12)-\frac{10}{3} (x-12)][/tex]= 0
We can again take common as (x-12).
⇒ [tex]x(x-12)[x-\frac{10}{3} ][/tex]=0
∴[tex](x-\frac{10}{3} ) (x-12)= 0[/tex]
We have two value for x, which is [tex]12 and \frac{10}{3}[/tex]
12 is invalid as, w= [tex](16-2x)= 16-2\times 12[/tex]
∴ 24 inches can not be cut out of 16 inches width.
Hence, the cut out size from cardboard is [tex]\frac{10}{3}\ inches [/tex]
Now, subtituting the value of x to find volume of the box.
Volume(v)= [tex](30-2x)\times (16-2x)\times x[/tex]
⇒ Volume(v)= [tex](30-2\times \frac{10}{3} )\times (16-2\times \frac{10}{3})\times \frac{10}{3}[/tex]
⇒ Volume(v)= [tex](30-\frac{20}{3} ) (16-\frac{20}{3}) (\frac{10}{3} )[/tex]
∴ Volume(v)= 725.93 inches³
