A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is

\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }

. $A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 } . Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle. A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } } . The square of the area of the triangle is equal to ( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 } , and the substitution x ^ { 2 } = t will transform this expression into the quadratic function - \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 ) Since we want to find the maximum value of t, we will substitute the value t = 18 R ^ { 2 } = x ^ { 2 } into the equation. Solving for t gives us the following minimum area of the shaded region: t = \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 } . B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } } . The square of the area of the triangle is equal to ( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 } The substitution x ^ { 2 } = t will transform this expression into the quadratic function - \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 ) Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is t = \frac { - b } { 2 a } = 18 R ^ { 2 } Substituting the value t = 18 R ^ { 2 } into the equation t = x ^ { 2 } gives us x ^ { 2 } = 18 R ^ { 2 } and consequently x = 3 R \sqrt { 2 } (The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to \frac { 9 \pi R ^ { 2 } } { 2 } - \frac { 1 } { 2 } x \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } } Substituting the value x = 3 R \sqrt { 2 } in the equation (2) gives us that the minimum value of the shaded region is equal to \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 } . C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to \frac { 1 } { 4 } 9 R ^ { 2 } x ^ { 2 } + \frac { 1 } { 4 } x ^ { 4 } . The substitution x ^ { 2 } = t will transform this into the quadratic function \frac { 1 } { 4 } 9 R ^ { 2 } t + \frac { 1 } { 4 } t ^ { 2 } Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is t = \frac { - b } { 2 a } = 9 R ^ { 2 } Substituting the value t = 9 R ^ { 2 } into the equation t = x ^ { 2 } gives us x ^ { 2 } = 9 R ^ { 2 } and consequently x = 3 R . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to \pi 9 R ^ { 2 } - \frac { 1 } { 2 } x \sqrt { 6 R ^ { 2 } - x ^ { 2 } } Substituting the value x = 3 R into the equation (2), we find that the minimum value of the shaded region is equal to \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 } . D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } } , which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as: x ^ { 2 } = \frac { - b } { 2 a } = 18 R ^ { 2 } We can then write x ^ { 2 } = 18 R ^ { 2 } as x ^ { 2 } = 18 R ^ { 2 } and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area: \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 } . E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to \frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 } . The substitution x ^ { 2 } = t will transform this into the quadratic function \frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } t - \frac { 1 } { 4 } t ^ { 2 } Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is t = \frac { - b } { 2 a } = 3 R ^ { 2 } Substituting the value t = 3 R ^ { 2 } into the equation t = x ^ { 2 } gives us x ^ { 2 } = 3 R ^ { 2 } and consequently x = R \sqrt { 7 } . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to \frac { \pi 9 R ^ { 2 } } { 4 } - \frac { 1 } { 2 } x \sqrt { 9 R ^ { 2 } - x ^ { 2 } } Substituting the value x = R \sqrt { 7 } into the equation (2), we find that the minimum value of the shaded region is equal to \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 } .$ Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.
A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to

A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }

.
The square of the area of the triangle is equal to

( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }

, and the substitution

x ^ { 2 } = t

will transform this expression into the quadratic function

- \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )

Since we want to find the maximum value of t, we will substitute the value

t = 18 R ^ { 2 } = x ^ { 2 }

into the equation. Solving for t gives us the following minimum area of the shaded region:

t = \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }

.
B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to

A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }

.
The square of the area of the triangle is equal to

( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }

The substitution

x ^ { 2 } = t

will transform this expression into the quadratic function

- \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )

Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is

t = \frac { - b } { 2 a } = 18 R ^ { 2 }

Substituting the value

t = 18 R ^ { 2 }

into the equation

t = x ^ { 2 }

gives us

x ^ { 2 } = 18 R ^ { 2 }

and consequently

x = 3 R \sqrt { 2 }

(The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to

\frac { 9 \pi R ^ { 2 } } { 2 } - \frac { 1 } { 2 } x \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }

Substituting the value

x = 3 R \sqrt { 2 }

in the equation (2) gives us that the minimum value of the shaded region is equal to

\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }

.
C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to

\frac { 1 } { 4 } 9 R ^ { 2 } x ^ { 2 } + \frac { 1 } { 4 } x ^ { 4 }

.
The substitution

x ^ { 2 } = t

will transform this into the quadratic function

\frac { 1 } { 4 } 9 R ^ { 2 } t + \frac { 1 } { 4 } t ^ { 2 }

Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is

t = \frac { - b } { 2 a } = 9 R ^ { 2 }

Substituting the value

t = 9 R ^ { 2 }

into the equation

t = x ^ { 2 }

gives us

x ^ { 2 } = 9 R ^ { 2 }

and consequently

x = 3 R

. With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to

\pi 9 R ^ { 2 } - \frac { 1 } { 2 } x \sqrt { 6 R ^ { 2 } - x ^ { 2 } }

Substituting the value

x = 3 R

into the equation (2), we find that the minimum value of the shaded region is equal to

\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }

.
D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle.
The square of the area of the triangle is equal to

A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }

, which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as:

x ^ { 2 } = \frac { - b } { 2 a } = 18 R ^ { 2 }

We can then write

x ^ { 2 } = 18 R ^ { 2 }

x ^ { 2 } = 18 R ^ { 2 }

and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area:

\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }

.
E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to

\frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }

.
The substitution

x ^ { 2 } = t

will transform this into the quadratic function

\frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } t - \frac { 1 } { 4 } t ^ { 2 }

Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is

t = \frac { - b } { 2 a } = 3 R ^ { 2 }

Substituting the value

t = 3 R ^ { 2 }

into the equation

t = x ^ { 2 }

gives us

x ^ { 2 } = 3 R ^ { 2 }

and consequently

x = R \sqrt { 7 }

. With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to

\frac { \pi 9 R ^ { 2 } } { 4 } - \frac { 1 } { 2 } x \sqrt { 9 R ^ { 2 } - x ^ { 2 } }

Substituting the value

x = R \sqrt { 7 }

into the equation (2), we find that the minimum value of the shaded region is equal to

\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }

Determine the Input That Produces the Largest or Smallest Output t=4916t = \frac { 49 } { 16 }t=1649​

Multiple Choice

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Determine the Input That Produces the Largest or Smallest Output $t = \frac { 49 } { 16 }$

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