c. 60/(sqrt(3)) in, 60/(sqrt(3)) in
To find the dimensions of the strongest beam, we need to maximize the product of width and the square of depth. Let w be the width and d be the depth of the rectangle.
From the figure, the diagonal of the rectangle is equal to the diameter of the cylinder, which is 2r = 60 inches. Using the Pythagorean theorem, we have:
(d^2 + w^2) = (2r)^2
(d^2 + w^2) = 3600
d^2 = 3600 - w^2
The strength of the beam is proportional to wd^2. So we need to maximize wd^2. Substituting the value of d^2 from the first equation, we get:
w(3600 - w^2) = 3600w - w^3
To find the maximum of this function, we take its derivative and set it equal to zero:
d/dw (3600w - w^3) = 3600 - 3w^2 = 0
3w^2 = 3600
w^2 = 1200
w = sqrt(1200) = 20*sqrt(3) inches
Substitute this back into the equation d^2 = 3600 - w^2:
d^2 = 3600 - (20*sqrt(3))^2
d^2 = 3600 - 1200
d^2 = 2400
d = sqrt(2400) = 20*sqrt(6) inches
Therefore, the dimensions of the strongest beam that can be cut from the cylindrical log are 60/(sqrt(3)) inches by 60/(sqrt(3)) inches.
A rectangular beam will be cut from a cylindrical log of radius r =30 inches. Suppose that the strength of a rectangular beam is proportional to the product of its width and the square of its depth. Find the dimensions of the strongest beam that can be cut from the cylindrical log.
The figure shows the cross-section of a cylindrical log of radius r and an inscribed rectangle in it. The sided of the rectangle are labeled as width and depth.
a.
Sqrt(3) in, 60 in
b.
60 in, sqrt(2 over 3) in
c.
60 divided by (sqrt(3)) in, 60 divided by (sqrt(3)) in
d.
60 (sqrt(2 over 3)) in, 60 (sqrt(2 over 3)) in
e.
60 divided by (sqrt(3)) in, 60 (sqrt(2 over 3)) in
1 answer
2 answers