Question

a rectangular sheet of metal 80 cm wide and 1000cm long is folded along the center to form a triangular trough and then the ends are capped. How deep should the trough be to maximize its capacity? Note: the volume is the area of a cross section times the length

Answer 1

Answer:

H=28.3 cm

Step-by-step explanation:

The shape of a vertical trough and its cross-section are shown in the images below. Note the width of the rectangular sheet of metal (80 cm) is folded into two equal sides of 40 cm each which form the hypotenuse of the triangle where the dimensions of the shape will be determined.

As indicated in the question, the volume of the trough is its area times the length:

$V=A_s\cdot L$

The cross-section is a triangle whose area can be computed as

$\displaystyle A_s=\frac{1}{2}\ X.H$

The sides H and X are related by using the Pythagora's formula

$X^2+H^2=40^2=1600$

Solving for X

$X=\sqrt{1600-H^2}$

The area is now

$\displaystyle A_s=\frac{1}{2}\ H\sqrt{1600-H^2}$

And the volume is

$\displaystyle V=\frac{1}{2}L.\ H\sqrt{1600-H^2}$

Since L=1000

$V=500H\sqrt{1600-H^2}$

To find the maximum volume, we compute the first derivative

$V'=500(H'\sqrt{1600-H^2}+H[\sqrt{1600-H^2}]'$

$\displaystyle V'=500\left(\sqrt{1600-H^2}-\frac{H^2}{\sqrt{1600-H^2}}\right)$

Equating V'=0

$\displaystyle \sqrt{1600-H^2}=\frac{H^2}{\sqrt{1600-H^2}}$

Operating

$1600-H^2=H^2$

Solving for H

$H=\sqrt{1600/2}=28.3\ cm$