Respuesta :
Answer:
The value of double derivative at x=4.834 is negative, therefore the trough have a maximum volume at x=4.834 inches.
Step-by-step explanation:
The dimensions of given metal strip are
Length = 160 inch
Width = 20 inch
Let the side bend x inch from each sides to make a open box.
Dimensions of the box are
Length = 160-2x inch
Breadth = 20-2x inch
Height = x inch
The volume of a cuboid is
[tex]V=length\times breadth \times height[/tex]
Volume of box is
[tex]V(x)=(160-2x)\times (20-2x)\times x[/tex]
[tex]V(x)=(160-2x)(20-2x)x[/tex]
[tex]V(x)=4 x^3 - 360 x^2 + 3200 x[/tex]
Differentiate with respect to x.
[tex]V'(x)=12x^2 - 720 x + 3200[/tex]
Equate V'(x)=0, to find the critical points.
[tex]0=12x^2 - 720 x + 3200[/tex]
Using quadratic formula,
[tex]x=30\pm 10\sqrt{\frac{\left(19\right)}{3}}[/tex]
The critical values are
[tex]x_1=30+10\sqrt{\frac{\left(19\right)}{3}}\approx 55.166[/tex]
[tex]x_2=30-10\sqrt{\frac{\left(19\right)}{3}}\approx 4.834[/tex]
Differentiate V'(x) with respect to x.
[tex]V'(x)=24x - 720[/tex]
The value of double derivative at critical points are
[tex]V'(55.166)=24(55.166) - 720=603.984[/tex]
[tex]V'(4.834)=24(4.834) - 720=-603.984[/tex]
Since the value of double derivative at x=4.834 is negative, therefore the trough have a maximum volume at x=4.834 inches.
Answer:
The question above is right!
Step-by-step explanation:
I double checked and took the test its all correct!
