Keith M. answered • 06/21/15
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B.S. in Applied Mathematics with 5+ years experience tutoring math
Hi Jon,
This is a good problem to test your understanding of single-variable calculus as it applies to basic optimization problems. To solve a problem like this one, we must carefully read the description and identify the constraints (if any) that apply as well as the optimization function, the thing we are trying to maximize or minimize.
Part a.
Here, we are looking for a rectangle whose base is on the x-axis between 0 and 1 (0 ≤ x ≤ 1), and one corner of the rectangle is on the curve y=x3. I will assume for the sake of non-triviality that this corner is the upper left-hand one, since the rectangle {(0, 0), (1, 1)} is the (trivial) solution to the upper right-hand case, {(0,0),(0,0)} is the (vacuous) solution for the lower right-hand case, and the lower left-hand case is unbounded (i.e. {(0, 0). (1, y)} is a solution for arbitrarily large y). Assume for simplicity's sake that the lower right-hand corner is at (1, 0), since this will provide maximal area.
Here is the problem framed in terms of its constraints and optimization function:
Find a point (x, y) = (x, x3). 0 ≤ x ≤ 1, maximizing f(x) = |1 - x|⋅|0 - y|.
Plugging in the constraints (x, y) = (x, x3) and 0 ≤ x ≤ 1 lets us simplify our goal, so we write:
maximize f(x) = x3 - x4 on 0 ≤ x ≤ 1.
This is a function of one variable, which can be optimized using straightforward calculus techniques.
Compute the derivative of the function and find its roots (known as the function's "critical points"):
f(x) = x3 - x4
f'(x) = 3⋅x2 - 4⋅x3
f'(x) = 0 = x2⋅(3 - 4⋅x)
∴ x = 0 or x = ¾.
Both of these critical points lie on our interval (0 ≤ x ≤ 1), and we must consider each of them as well as the endpoints. x = 0 and x = 1 are both minimum values, since f(0) = 0 and f(1) = 0, but x = ¾ is the maximum value, which can be verified by checking that the sign of the derivative switches from positive to negative at this point.
The rectangle has width 1 - ¾ = ¼ and height ¾3. The area of the rectangle is then ¼⋅¾3 = 27/256.
Part b.
The extra constant coefficient here doesn't change much except the value of the area. In fact, the same analysis above can be reused with a C tacked onto each equation involving f and f'.
The answer is similarly just scaled by C:
The rectangle has width 1 - ¾ = ¼ and height C⋅¾3. The area of the rectangle is then C⋅¼⋅¾3 = (27/256)⋅C.
Part c.
Don't let the fact that part b introduces a variable named C and part c uses a variable named B throw you off! There's not too much else that's tricky here.
Since the problem is a little bit different this time, I'll start by writing the constraints and optimization function again.
Find a point (x, y) = (x, C⋅x3). 0 ≤ x ≤ B, maximizing f(x) = |B - x|⋅|0 - y|.
Notice that the lower right-hand corner of the rectangle is now (B, 0), meaning our optimization function has changed as well as the constraint on x (0 ≤ x ≤ B).
Just as before, we can use the constraints to rewrite the goal function:
maximize f(x) = C⋅B⋅x3 - C⋅x4 on 0 ≤ x ≤ B.
Compute the derivative f'(x) and obtain its roots to find the critical points for f:
f(x) = C⋅B⋅x3 - C⋅x4
f'(x) = 3⋅C⋅B⋅x2 - 4⋅C⋅x3
f'(x) = 0 = C⋅x2⋅(3⋅B - 4⋅x)
∴ x = 0 or x = ¾⋅B.
Again as before, we notice that on the interval 0 ≤ x ≤ B f(0) and f(B) are local minimums for f, and f(¾⋅B) is a local maximum.
Using the point (¾⋅B, ¾3⋅C⋅B3), as the upper left-hand corner, we find a rectangle with width ¼⋅B and height ¾3⋅C⋅B3, yielding a maximum area of ¼⋅¾3⋅C⋅B4.
Part d.
Again, the constraints and optimization function are similar to part c, but the point corresponding to the upper left-hand corner of our rectangle is now (x, C⋅xn).
We proceed as before, using the constraints to simplify the optimization function:
maximize f(x) = C⋅B⋅xn - C⋅xn+1 on 0 ≤ x ≤ B.
Solve the optimization problem by finding the critical points:
f(x) = C⋅B⋅xn - C⋅xn+1 on 0 ≤ x ≤ B.
f'(x) = n⋅C⋅B⋅xn-1 - (n+1)⋅C⋅xn
f'(x) = 0 = C⋅xn-1⋅(n⋅B - (n+1)⋅x)
∴ x = 0 or x = B⋅n/(n+1).
We can conclude in the same fashion as before that the upper left-hand corner should be placed at (B⋅n/(n+1), C⋅Bn⋅[n/(n+1)]n). This gives a rectangle with width B⋅[1/(n+1)] and height C⋅Bn⋅[n/(n+1)]n, giving an area of C⋅Bn+1⋅nn⋅[1/(n+1)]n+1.
Notice that each of these generalizations yield the answers we had found for the previous parts when the correct values of n, B, and C (i.e. n=3, B=1, C=1) are plugged in.
