# Isaiah's Math Lessons : Maxima & Minima

• How to find the length of the shortest segment tangent to an ellipse

Formula:

L = a + b

Where a and b are the semi-major and semi-minor axis of an ellipse

Example: Find the length of the shortest tangent to x2/9 + y2/4 = 1

Solution:

If we will apply calculus here, it would be a lengthy solution. Thus, we just need to use the formula above.

a= √9 = 3, b = √4 = 2.

Length = a + b = 3 + 2 = 5 units

How to find two numbers given that their sum is S and product of x a  and y b is to be maximized.

Formula:

X = S * a / ( a + b )

Y = S * a / ( a + b)

Example: Find 2 positive numbers such that their sum is 20 and the product of the cube of the first and the square of the second number is to be maximized.

Solution:

Product = x3 y2

Applying the formula above, we will get the following values below:

X = 20 * 3 / ( 3 + 2 ) = 12

Y = 20 * 2 ( 3 + 2 ) = 8

The numbers required are 12 and 8.

Maximum Area of a right triangle given the hypotenuse H.

Formula:

Area max = H2 / 4

Example: Find the maximum area of a triangle with hypotenuse of length 8 meters.

Solution:

Applying the formula:

Areamax = 82/4

Thus, maximum area is equal to 16 square meters.

Maximum Area of an isosceles triangle with perimeter P.

Formula:

Areamax = √3 * P/ 36

Example: What is the maximum area of an isosceles triangle with perimeter of 12 inches?

Solution:

Area = √3 * 12/ 36

Area = 4√3 in2

Maximum volume of the cylinder that can be inscribed in a cone of radius R and height H.

Formula:

Volumemax = 4 VolumeCone / 9

r = 2R/3

h = H/3

Where r and h are the radius and height of the cylinder respectively

Example: What is the volume of the largest cylinder that can be placed inside a cone of radius 3 cm and height of 6 cm?

Solution:

Vcone = Pi x r2 x h / 3

Substitute r = 3 and h = 6

V = 18pi cubic cm.

Vcylinder =  = 4 VolumeCone / 9 = 4 x 18 Pi / 9

V = 8pi cubic cm.

Area of the maximum rectangle that can be inscribed in a semi-circle of radius R.

Formula:

Area = R2

Example: What is the area of the largest rectangle that can be inscribed inside a semi-circle of diameter 20 cm?

Solution:

Area = R

R = D/2 = 10cm

Area = 102 = 100 square cm.

Minimum volume of a circular cone that can be circumscribed about a sphere of radius R.

Formula:

Volume = 8 Pi x r3 / 3

Example: Find the volume of the smallest circular cone that can be circumscribed about a sphere of radius 3 cm.

Solution:

Volume =   8 Pi x 33 / 3

Volume = 72pi cubic cm.

Volume of the largest right pyramid with square base that can be inscribed in a sphere of radius R.

Formula:

Volume = 64 R3 / 81

Example: Find the largest right pyramid with a square base that can be inscribed in a sphere of radius 6 meters.

Solution:

Volume = 64 x 63 / 81

Volume = 512/3 cubic meters

Longest length of girder of negligible width that can pass round two intersecting corridors

Formula:

Length = (a2/3 + b2/3 )3/2

Example: A steel girder of length X feet long is moved on rollers along a passage way 27 feet wide and into a corridor at right angles to the passageway. What is the value of X if the width of the girder is negligible and the narrowest corridor is 64 feet wide?

Solution:

X=  (272/3 + 642/3 )3/2

X = 125 feet.

Number Problems Technique.

Example: What number exceeds its square by the maximum amount?

Solution:

Let x – number

x2 - square of the number

D = x - x2

Applying 1st derivative then equate to 0

1 – 2x = 0

x = ½ (this will give the maximum amount when subtracted by its square)

Minimum amount of Post problem.

Formula:

Distance = X * a/ (a + b)

Where X, a and b be the distance of the 2 posts, height of the 1st post and height of the 2nd post respectively.

Example: Two post, one 6 meters and the other 12 meters high are 15 meters apart. If the posts are supported by a cable running from the top of the first post to a stake on the ground and then to the top of the second post, find the distance from the lower post to the stake to use the minimum amount of wire.

Solution:

Distance = 15 *  6 / ( 6 + 9 )

Distance = 6 meters

Formula: Example: Find the radius of curvature at any point on the curve y = - ln (cosx).

Solution:

y’ = d (-ln cosx) = tan x

y’’ = d (tanx) = sec2x Area of the largest rectangle that can be inscribed in any right triangle with legs a and b such that the sides of the rectangle are parallel to the legs of the right triangle.

Formula:

Area = ¼ ab

Example: What is the area of the largest rectangle that can be inscribed in a right triangular lot with legs 6 meters and 8 meters long?

Solution:

Area = ¼ * 6*8

Area = 12 square meters

Maximum angle subtended by a hanging picture frame. Formula:

D = √(XY)

Example: A statue 3 meters high is standing on a base of 4 meters high. If an observer’s eye is 2 meters above the ground, how far should he stand from the base in order that the angle subtended by the statute is maximum?

Solution:

D =√(XY) where X = 4m – 2m = 2m and Y = 2m + 3m = 5m

D = √(2x5)

D = √10 meters

In how many equal parts should you divide 100 so that the continued product of those equal parts will be maximum?

Formula:

N =floor (Sum/e)

where e is the Euler’s constant approximately equal to 2.71828

Floor function means the greatest integer less than or equal to

Solution:

N = floor (100/e) ~ 36

N = 36

1

14

1

1

48

6

1

1