Respuesta :
Answer:
The volume in such a package is 10,415.41 in³
Step-by-step explanation:
Consider the provided information.
A parcel delivery service will deliver a package only if the length plus the girth (distance around, taken perpendicular to the length) does not exceed 104 inches.
Let the dimension are x by x by y.
Where x is the variable for the square base package and y is the variable for length.
Thus l=x, b=x and h=y
Then the volume of the box is: [tex]V(x)=x^2y[/tex] (∵V=lbh)
The maximum combined length and girth is 104.
Therefore, [tex]4x+y=104[/tex]
[tex]y=104-4x[/tex]
Substitute the value of y in volume of the box.
[tex]V(x)=x^2(104-4x)[/tex]
[tex]V(x)=104x^2-4x^3[/tex]
[tex]V'(x)=208x-12x^2[/tex]
Substitute V'(x)=0.
[tex]0=208x-12x^2[/tex]
[tex]-4x(3x-52)=0[/tex]
[tex]x=0\ or\ x=\frac{52}{3}[/tex]
Now apply second derivative test.
[tex]V''(x)=208-24x[/tex]
[tex]V''(0)=208-24(0)>0[/tex] (Min)
[tex]V''(\frac{52}{3})=208-24(\frac{52}{3})<0[/tex] (Max)
If x=52/3 then [tex]y=104-4(\frac{52}{3})=\frac{104}{3}[/tex]
Substitute x = 52/3 and y = 104/3 in [tex]V(x)=x^2y[/tex]
[tex]V(x)=(\frac{52}{3})^2\times \frac{104}{3}=10,415.41[/tex]
Hence, the volume in such a package is 10,415.41 in³
