Mathematics: A160-inch strip of metal 20 inches wide is to be made into a small open trough by bending up two sides on the long side, at right angles to the base. the sides will be the same height, x. if the trough is to have a maximum volume, how many inches should be turned up on each side?

3 rows with 15 columns or it could be 15 rows with 3 columns. Hope it helped ;)...

A160-inch strip of metal 20 inches wide is to

annaebrown9737

30.12.2021

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

$V=length\times breadth \times height$

Volume of box is

$V(x)=(160-2x)\times (20-2x)\times x$

V(x)=(160-2x)(20-2x)x

V(x)=4 x^3 - 360 x^2 + 3200 x

Differentiate with respect to x.

V'(x)=12x^2 - 720 x + 3200

Equate V'(x)=0, to find the critical points.

0=12x^2 - 720 x + 3200

Using quadratic formula,

$x=30\pm 10\sqrt{\frac{\left(19\right)}{3}}$

The critical values are

$x_1=30+10\sqrt{\frac{\left(19\right)}{3}}\approx 55.166$

$x_2=30-10\sqrt{\frac{\left(19\right)}{3}}\approx 4.834$

Differentiate V'(x) with respect to x.

V'(x)=24x - 720

The value of double derivative at critical points are

V'(55.166)=24(55.166) - 720=603.984

V'(4.834)=24(4.834) - 720=-603.984

Since the value of double derivative at x=4.834 is negative, therefore the trough have a maximum volume at x=4.834 inches.

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