Is there somebody that could help me with this problem please? Thanks in advance!

Question

You are given a rectangular piece of paper that has length x=13.8 cm and height y= 12 cm. The lower right corner is to be folded to the top edge forming a triangle. Determine the maximum and minimum area of a triangle that can be constructed. 

Answer 1

Referring to your picture, let's give names to the various points.

Let the shaded triangle of interest be ABC, where

A is the vertex along the bottom edge,

B is the vertex along the right-side edge,

C is the vertex along the top edge.

Let

U be be the upper right corner of the paper,

L be the lower right corner of the paper,

M be the lower left corner of the paper.

From the point A draw a line parallel to the side y;

let the intersection of this line with the upper edge be called D.

Now let's give names to the lengths of the various line segments.

Let

a be the side of the triangle ABC opposite A, i.e., side BC,

b be the side of the triangle ABC opposite B, i.e., side AC,

v be the length of UC.

Our independent variable will be v, and we want to find values of v that

maximize/minimize the area of the triangle ABC,

which is ab/2.

Note some identities among the lengths of various line segments:

LA = b

LB = a

UB = y-a

UD = b

CD = b-v

We now have all the lengths in the picture expressed in terms of

the known x,y and the three unknown a,b,v.

We will proceed in the following steps.

Step 1: Express the area f of the triangle ABC in terms of the independent variable v, not

in terms of a,b.

Step 2: Calculate the range of the function f(v)

Step 3: Calculate any critical points of f(v)

Step4: Minimum and Maximum is either at a critical point or at boundary of the domain.

-----------------

Step 1:

We can express a in terms of v by using the Pythagorean theorem applied to the triangle BCU:

v² + (y-a)² = a²

a = (v² + y²)/2y

We can express b in terms of v by using the Pythagorean theorem applied to the triangle ADC:

(b-v)² + y² = b²

b = (v² + y²)/2v

The area of the triangle ABC is

f(v) = ab/2 = (v² + y²)²/8vy

Note that f(v) is independent of x. The quantity x plays a role only in determining the domain of f(v).

-----------------

Step 2:

For the above derivation to make sense geometrically

we have to assume that all the sides are non-negative:

Constraint 0: UC ≥ 0, i.e., v ≥ 0

Constraint 1: BC ≥ 0, i.e., a ≥ 0

(v² + y²)/2y ≥ 0

That is always satisfied.

Constraint 2: AC ≥ 0, i.e., b ≥ 0

(v² + y²)/2v ≥ 0

That is implied by Constraint 0.

Constraint 3: UB ≥ 0, i.e., y-a ≥ 0

y - (v² + y²)/2y ≥ 0

2y² - v² - y² ≥ 0

v ≤ y

Constraint 4: CD ≥ 0, i.e. b-v ≥ 0

(v² + y²)/2v - v ≥ 0

v ≤ y

Constraint 5: AM ≥ 0, i.e. x-b ≥ 0

x - (v² + y²)/2v ≥ 0

v² - 2xv + y² ≤ 0

For that to be satisfied, v must lie between the two roots

v = (2x ± √(4x² - 4y²))/2 = x ± √(x² - y²)

From the above constraints we collect those that are not implied by others and

substitute actual numbers.

Constraint 0: v ≥ 0

Constraint 3: v ≤ 14

Constraint 5: 9.28 ≤ v ≤ 21.12

As the intersection of the above constraints, the domain of the function f(v) is the interval [9.28, 14].

-----------------

Step 3:

Critical points occurs where f'(v) = 0:

2(v² + y²)2v/8vy - (v² + y²)²/8v²y = 0

2(v² + y²)2v² - (v² + y²)² = 0

4v² - v² - y² = 0

v = y/√3

Substituting actual numbers, the critical point occurs at

v = 14/√3 = 8.08

As that lies outside of the domain, it is not a candidate for minimum or maximum.

-----------------

Step 4:

We have only two candidates for minimum and maximum -- the bounds of the interval.

f(9.28) = (9.28² + 14²)²/8×9.28×14 = 76.58

f(14) = (14² + 14²)²/8×14×14 = 98

The minimum is 76.58 and the maximum is 98.

The minimum is archived by making the crease through point M, and

the maximum is archived by making the crease through point U.