Can anyone help me with this question pleaseee
1 answer:
You might be interested in
Answer:
The dimension of the box is 18.66 in by 2.66 in by 1.17 in.
Step-by-step explanation:
Given that, The cardboard is 21 in. long and 5 in. wide.
Assume,x be the length of the each sides of the square .
Then the length of the box is = (21-2x) in.
The breadth of the box is =(5-2x) in.
The height of the box is = length of side of the square
= x in.
The volume of the box = length × wide × height
=(21-2x)(5-2x)x
=(105-52x+4x²)x
=()
Let
V=
Differentiating with respect to x
Again differentiating with respect to x
To find the dimension of the box, we set V'=0
Applying the quadratic formula , here a=12, b= -104 and c =105
=7.5, 1.17
For x= 7.5 , the wide of the box will be negative.
∴x=1.17 in.
Since at x= 1.17, V''<0 , therefore at x=1.17 in. the volume of the given box will be maximum.
The length of the box is = [21-(1.17×2)]=18.66 in.
The wide of the box is = [5-(1.17×2)]=2.66 in.
The height of the box is = 1.17 in.
The dimension of the box is 18.66 in by 2.66 in by 1.17 in.