Respuesta :
Answer:
The sum of dimensions is 12 in.
Step-by-step explanation:
Given that, the dimension of the cardboard is 8 in by 8 in.
Let the length of side of the squares which are cut out from cardboard be x
The height of the folded box is = x in
The length of the folded box is =(8-2x) in.
The width of the folded box is = (8-2x) in.
Then the volume of the folded box is= Length × width × height
=(8-2x)(8-2x)x cubic in
=(64x-32x²+4x³) cubic in
Let,
V=64x-32x²+4x³
Differentiating with respect to x
V'= 64 -64x+12x²
Again differentiating with respect to x
V''= -64+24x
For maximum(or for the minimum value)
V'=0
⇒64 -64x+12x²=0
⇒4(3x²-16x+16)=0
⇒3x²-16x+16=0
⇒3x²-12x-4x+16=0
⇒3x(x-4)-4(x-4)=0
⇒(3x-4)(x-4)=0
[tex]\Rightarrow x=\frac43, 4[/tex]
Now, [tex]V''|_{x=4}=-64+24\times 4=32 >0[/tex]
[tex]V''|_{x=\frac43}=-64+24\times \frac43=-32 <0[/tex]
Since at [tex]x=\frac43[/tex], V''<0, So at [tex]x=\frac43[/tex] , the volume of the box maximum
The height of the folded box is =[tex]\frac43[/tex] in
The length of the folded box is [tex]=(8-2.\frac43)[/tex]
[tex]=\frac{16}3[/tex] in
The width of the folded box is [tex]=(8-2.\frac43)[/tex]
[tex]=\frac{16}3[/tex] in
The sum of dimensions is
=Length+width+height
[tex]=(\frac{16}3+\frac{16}3+\frac43)[/tex] in
[tex]=(\frac{16+16+4}{3})[/tex] in
[tex]=\frac {36}3[/tex] in
=12 in
