MATH 100 V1A
November 12th – Practice problems
Hints and Solutions
1. What is the shortest distance between the curve y − x3/2 = 1 and the point (8, 1)?
Solution: Using the Pythagorean Theorem, we calculate the distance, D, between a
point (x, y) on the curve and the point (8, 1) to be
p
D = (x − 8)2 + (y − 1)2 .
However, if (x, y) is on the curve, then y = 1 + x3/2 , so this simplifies to a function of
x only:
p
D(x) = (x − 8)2 + x3 .
Note that the domain of the function f (x) = 1 + x3/2 is {x ∈ R | x ≥ 0}, which means
that the domain of the problem is also {x ∈ R | x ≥ 0}.
Taking the derivative of D with respect to x we get:
1
1
· (2(x − 8) + 3x2 ) = p
· (2x − 16 + 3x2 ).
D0 (x) = p
2
3
2
3
2 (x − 8) + x
2 (x − 8) + x
Now, the denominator is always bigger than zero when x ≥ 0, so to find the critical
points of D, we just need to solve for when D0 (x) = 0. This will occur when the
numerator is zero: 2x − 16 + 3x2 = 0. We can solve this equation using the quadratic
formula to get x = 2 and x = − 38 . Since − 38 is not in the domain of the problem, we
don’t need to consider it. To determine if x = 2 corresponds to a local maximum or a
local minimum, we make the following table:
0
D (x)
x ∈ (0, 2) x ∈ (2, ∞)
−
+
From this we see that x = 2 corresponds to a local minimum and, since D is continuous
on the domain of the problem [0, ∞), we can conclude that it corresponds to a global
minimum. Therefore √
the shortest distance between the curve y − x3.2 = 1 and the
point (8, 1) is D(2) = 44.
2. Of all isosceles triangles of a given perimeter, prove that the largest is equilateral.
Hint: Let P be the perimeter of the triangle and let x, x and P − 2x be the length of
the sides (so P − 2x is the length of the side which is possibly a different length than
the other two). If the side
q of length P − 2xqcorresponds to the base of the triangle,
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show that the height is x2 − ( P2 − x)2 = P x − P4 . Use this to find an equation
for the area of the triangle as a function of x and maximize this function.
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