Chapter 4, part 1 Review
1. Two numbers add to 16.
a. Express the product of the two numbers in terms of a single variable.
x+y=16, y = 16-x, xy = x(16-x) = 16x – x2
b. Express the sum of the squares of the two numbers in terms of single variable.
x2 + y2 = x2 + (16-x)2 = 2x2 – 32x + 256
2. The total surface area of a right circular cylinder is 14 sq inches. Express the volume as a function
of the radius. The surface area of a cylinder is 2rh, where h is the height, and the volume is r2h.
14 = 2rh, h= 7/r
V = r2h = r2(7/r) = 7r
3. A line is drawn from the origin O to a point P(x,y) in the first quadrant on the graph of y =1/x. A
line is drawn through P perpendicular to the x-axis, meeting the x axis at B.
a. Draw a picture of the situation
b. Derive the perimeter of triangle OPB as a function of x
c. Try to express the area of OPB as a function of x
Perimeter is OP + PB + OB, PB= y, OB = x, OP = √𝑥 2 + 𝑦 2
From the curve, point P is on y=1/x, so y = 1/x
1
Perimeter = 1/x + x + √𝑥 2 + (𝑥)2
A = ½ xy = 1/2
4. A wire of length of L (a constant) is cut into two pieces. The first is bent into a square, the second
into an equilateral triangle. Express the total area of the square and triangle as a function of x,
where x is the length of wire used for the triangle.
perimeter of triangle = x = 3y, where y is the length of a side. The area of a equilateral triangle is
its base times it height. Its base is y = x/3, it height is y sin 60= √3y/2.
A =(1/2) (x/3)(x/3)√3 /2 = x2√3/36
The area of square is s2, where s is the length of a side. The perimeter of the square is 4s = L-x, or
s = (L-x)/4. Area is (L-x)2/16
Total area is (L-x)2/16 + x2√3/36
5. The base of a rectangle lies on the x axis while the upper two vertices lie on the parabola
y = 10 – x2. If the upper right vertex of the rectangle has coordinates (x,y), what is the area of the
rectangle as a function of x?
If one side, the width, is on the x axis, to get a rectangle, the height, y, has to be the length of the
rectangle.The width is 2x, the height is y = 10-(x)2. Thus the area of the rectangle is
2x(10-(x)2) = 2x( 10-x2) = 20x – 2x3
6. Express the length AE in the figure below as a function of x. The angles with vertices at C and D
are right angles, AC has length x, CD has length 4, and CB has length 5.
Note that we have similar triangles. AD/AC = const, etc.
We need to find AB , and AB/AE is this same ratio
We have AC = x, and CD = 4 and CB = 5.
AB2 = AC2 + CB2 = x2 + 25 by Pythagoras. Or AB = √𝑥 2 + 25
AE/AD= AB/AC or AE = (AB)(AD)/(AC)
= √𝑥 2 + 25(x + 4) / x
E
B
A
C
D