1. How close does the semicircle x^2 + y^2 = 16 come close to the point (3^1/2, 1)? (Hint: if you minimize the square of the distance, you can avoid dealing with square roots.)
2. The U.S. Postal Service wil accept a box for domestic shipment only if the some of its length and girth (girth being total distance around the package once) does not exceed 92 inches. What dimensions will give a box with a square end the largest possible volume?
3. The Strength of a rectangular wooden beam is proportional to its width times the square of its depth. Find the dimensions of the strongest rectangular beam that can be cut from a 12-inch diameter log.
I am in the first semester of calculus in college, so this work would have to be done with that, can anyone help me? I don't understand this at all.
#3.
We are told the strength is proportional to the product of the width and the square of the height(depth).
So, we can write
S is the strength and k is a constant that depends on the wood type, etc.
We find a function that relates w and h. Draw a line from the center of the log to the corner of the beam.
We get a right triangle.
Now, sub into the strength equation:
Now, differentiate, set to 0 and solve for w. h will follow.
Let be the strength of the beam
Let be the width of the beam
Let be the depth of the beam
Let be the proportionality constant
Since the strength is proportional to the width times the square of the depth, we have that:
thus, to maximize the strength, we need to maximize the dimensions of the rectangle.
now, imagine the end of the log as a circle centered at the origin. the equation of this circle is therefore given by ....(i use w on the x-axis and d on the y-axis)
we want to maximize the rectangle we can get in the upper half of the circle. draw a diagram and you will realize that the length of the rectangle is given by and the height (or depth) is given by
thus we have the Area = height*width being
let's maximize this. our constraint is that
thus we have now find the values for and that maximizes this. (find and set it to zero and solve for , then plug the value you got for into the original equation to solve for )
(alternatively, which may be more appropriate for this question, you can maximize the strength directly by plugging in into the equation for . yes, i think that would be better)