Respuesta :
Answer:
The dimension of the box is 17.66 in by 7.66 in by 2.67 in.
Therefore the volume of the box is 361.19 [tex]in^3[/tex].
Step-by-step explanation:
Given that the dimensions of a cardboard is 23 in by 13 in.
Let the side of the square be x in.
Then the length of the box= (23-2x) in
and the width of the box =(13-2x) in
and height = x in.
The volume of the box is = length ×width × height
=[(23-2x)(13-2x)x] [tex]in^3[/tex]
=(299x-72x² +4[tex]x^3[/tex]) [tex]in^3[/tex]
∴V=299x-72x² +4x³
Differentiating with respect x
V'= 299-144x+12x²
Again differentiating with respect x
V''= -144+24x
To find the dimensions, we set V'=0
∴299-144x+12x²=0
Applying Sridharacharya formula that is the solution of a quadratic equation ax²+bx+c is [tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Here a=12, b=-144 , c=299
[tex]\therefore x=\frac{-(-144)\pm\sqrt{(-144)^2-4.12.299}}{2.12}[/tex]
[tex]\Rightarrow x= 9.33, 2.67[/tex]
If we take x=9.33 in, then the width of the box [13-(2×9.33)] will negative.
∴x = 2.67 in
If at x = 2.67, V''<0 , then the volume of the box will be maximum or V''>0 then volume of the box will be minimum.
[tex]V''|_{x=2.67}=-144+(24\times 2.67)=-79.92<0[/tex]
Therefore at x = 2.67, the volume of the box maximum.
The length of the box =[23-(2×2.67)] in
=17.66 in
The width of the box =[13-(2×2.67)] in
=7.66 in
The height of the box= 2.67 in
The dimension of the box is 17.66 in by 7.66 in by 2.67 in.
Therefore the volume of the box is =(17.66×7.66×2.67) [tex]in^3[/tex]
=361.19 [tex]in^3[/tex]
