Exercise 35 Page 600

The length of the rectangle is 6. Now we will find the width w of the rectangle. The width is equal to the y-coordinate of the upper vertices. To find it, we will substitute x= 3 into the equation of the function.

y=- 0.3x^2+4

Substitute

x= 3

y=- 0.3( 3^2)+4

▼

Simplify

CalcPow

Calculate power

y=- 0.3(9)+4

Multiply

Multiply

y=- 2.7+4

AddTerms

Add terms

y=1.3

The width of the rectangle is 1.3.

Recall the formula for the area of a rectangle. A=l w In our case l=6 and w=1.3. Let's calculate the area of the rectangle!

A=l w

SubstituteII

l= 6, w= 1.3

A=6(1.3)

Multiply

Multiply

A=7.8

The area of the rectangle is 7.8.

First, we will consider only the highlighted vertex. Let y be the side length of the square. It is the y-coordinate of the upper vertices as well. Since the y-axis cuts the square in half we receive the following equation. y/2=x ⇔ y=2x The vertex must be also included in the parabola. Using this information we can write the second equation. y=- 0.3x^2+4 Let's write our system of equations. y=2x & (I) y=- 0.3x^2+4 & (II) We will solve this system using the Substitution Method. Note that the vertex which we are considering lays in the first quadrant. Therefore, we will only consider the coordinate pairs (x,y) where x>0 and y>0 as our solutions.

y=2x & (I) y=- 0.3x^2+4 & (II)

Substitute

(II):y= 2x

y=2x 2x=- 0.3x^2+4

SubEqn

(II):LHS-2x=RHS-2x

y=2x 0=- 0.3x^2-2x+4

Let's solve the second equation using the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a In our case a=- 0.3, b=- 2, and c=4. Let's substitute these values into the formula.

x=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

x=- (- 2)±sqrt((- 2)^2-4(- 0.3)(4))/2(- 0.3)

▼

Simplify

NegNeg

- (- a)=a

x=2±sqrt((- 2)^2-4(- 0.3)(4))/2(- 0.3)

NegBaseToPosBase

(- a)^2=a^2

x=2±sqrt(4-4(- 0.3)(4))/2(- 0.3)

MultNegPos

(- a)b = - ab

x=2±sqrt(4-4(- 1.2))/2(- 0.3)

MultNegNegOnePar

- a(- b)=a* b

x=2±sqrt(4+4.8)/2(- 0.3)

AddTerms

Add terms

x=2±sqrt(8.8)/2(- 0.3)

UseCalc

Use a calculator

x=2± 2.966.../2(- 0.3)

MultPosNeg

a(- b)=- a * b

x=2± 2.966.../- 0.6

Let's split the result into the positive and negative cases.

x=2± 2.966.../- 0.6
x=2+2.996.../- 0.6	x=2-2.966.../- 0.6
x=- 8.277...	x=1.610...
x≈ - 8.28	x≈ 1.61

Recall that we only consider the positive solutions. Therefore, our only solution is x≈ 1.61. Using the first equation from the system, y=2x, we can find the value of y.

y=2x

x ≈ 1.61

y≈ 2( 1.61)

Multiply

Multiply

y≈ 3.22

The highlighted vertex of the square is the point (1.61,3.22). This information allows us to determine the coordinates of the remaining three vertices.

The vertices of the square are (1.61,0), (1.61,3.22), (- 1.61,3.22), and (- 1.61,0).

A=s^2 Here, s is the side length of the square. In our case s= 3.22. Let's substitute this value into the formula and simplify!

A=s^2

Substitute

s= 3.22

A= 3.22^2

UseCalc

Use a calculator

A=10.3684

RoundDec

Round to 2 decimal place(s)

A≈ 10.37

The area of the square is about 10.37.