Answer :
To find the largest possible value of [tex]\( x \)[/tex] for the given problem, we start by applying the Pythagorean theorem to the cone. The theorem relates the slant height [tex]\( l \)[/tex], the radius [tex]\( r \)[/tex], and the height [tex]\( h \)[/tex] of the cone through the equation:
[tex]\[ l^2 = r^2 + h^2 \][/tex]
Given:
- Radius ([tex]\( r \)[/tex]) = [tex]\( x + 3 \)[/tex]
- Height ([tex]\( h \)[/tex]) = [tex]\( 5x - 3 \)[/tex]
- Slant height ([tex]\( l \)[/tex]) = [tex]\( \sqrt{15} \)[/tex]
We substitute these expressions into the Pythagorean theorem:
[tex]\[ (\sqrt{15})^2 = (x + 3)^2 + (5x - 3)^2 \][/tex]
First, we simplify the left side of the equation:
[tex]\[ 15 = (x + 3)^2 + (5x - 3)^2 \][/tex]
Next, we expand both squared terms on the right side:
[tex]\[ (x + 3)^2 = x^2 + 6x + 9 \][/tex]
[tex]\[ (5x - 3)^2 = 25x^2 - 30x + 9 \][/tex]
Now, add these two expressions:
[tex]\[ 15 = x^2 + 6x + 9 + 25x^2 - 30x + 9 \][/tex]
[tex]\[ 15 = x^2 + 25x^2 + 6x - 30x + 9 + 9 \][/tex]
[tex]\[ 15 = 26x^2 - 24x + 18 \][/tex]
To solve for [tex]\( x \)[/tex], we first move all terms to one side of the equation to set it to zero:
[tex]\[ 26x^2 - 24x + 18 - 15 = 0 \][/tex]
[tex]\[ 26x^2 - 24x + 3 = 0 \][/tex]
This is a quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 26 \)[/tex], [tex]\( b = -24 \)[/tex], and [tex]\( c = 3 \)[/tex].
We solve the quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the formula:
[tex]\[ x = \frac{-(-24) \pm \sqrt{(-24)^2 - 4 \cdot 26 \cdot 3}}{2 \cdot 26} \][/tex]
[tex]\[ x = \frac{24 \pm \sqrt{576 - 312}}{52} \][/tex]
[tex]\[ x = \frac{24 \pm \sqrt{264}}{52} \][/tex]
[tex]\[ x = \frac{24 \pm 2\sqrt{66}}{52} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{66}}{26} \][/tex]
Thus, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \frac{12 - \sqrt{66}}{26} \][/tex]
[tex]\[ x = \frac{12 + \sqrt{66}}{26} \][/tex]
To find the largest possible value of [tex]\( x \)[/tex], we select the larger of the two solutions:
[tex]\[ x = \frac{12 + \sqrt{66}}{26} \][/tex]
So, the largest possible value of [tex]\( x \)[/tex] is [tex]\( \frac{12 + \sqrt{66}}{26} \)[/tex].
[tex]\[ l^2 = r^2 + h^2 \][/tex]
Given:
- Radius ([tex]\( r \)[/tex]) = [tex]\( x + 3 \)[/tex]
- Height ([tex]\( h \)[/tex]) = [tex]\( 5x - 3 \)[/tex]
- Slant height ([tex]\( l \)[/tex]) = [tex]\( \sqrt{15} \)[/tex]
We substitute these expressions into the Pythagorean theorem:
[tex]\[ (\sqrt{15})^2 = (x + 3)^2 + (5x - 3)^2 \][/tex]
First, we simplify the left side of the equation:
[tex]\[ 15 = (x + 3)^2 + (5x - 3)^2 \][/tex]
Next, we expand both squared terms on the right side:
[tex]\[ (x + 3)^2 = x^2 + 6x + 9 \][/tex]
[tex]\[ (5x - 3)^2 = 25x^2 - 30x + 9 \][/tex]
Now, add these two expressions:
[tex]\[ 15 = x^2 + 6x + 9 + 25x^2 - 30x + 9 \][/tex]
[tex]\[ 15 = x^2 + 25x^2 + 6x - 30x + 9 + 9 \][/tex]
[tex]\[ 15 = 26x^2 - 24x + 18 \][/tex]
To solve for [tex]\( x \)[/tex], we first move all terms to one side of the equation to set it to zero:
[tex]\[ 26x^2 - 24x + 18 - 15 = 0 \][/tex]
[tex]\[ 26x^2 - 24x + 3 = 0 \][/tex]
This is a quadratic equation in standard form [tex]\( ax^2 + bx + c = 0 \)[/tex], where [tex]\( a = 26 \)[/tex], [tex]\( b = -24 \)[/tex], and [tex]\( c = 3 \)[/tex].
We solve the quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the formula:
[tex]\[ x = \frac{-(-24) \pm \sqrt{(-24)^2 - 4 \cdot 26 \cdot 3}}{2 \cdot 26} \][/tex]
[tex]\[ x = \frac{24 \pm \sqrt{576 - 312}}{52} \][/tex]
[tex]\[ x = \frac{24 \pm \sqrt{264}}{52} \][/tex]
[tex]\[ x = \frac{24 \pm 2\sqrt{66}}{52} \][/tex]
[tex]\[ x = \frac{12 \pm \sqrt{66}}{26} \][/tex]
Thus, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \frac{12 - \sqrt{66}}{26} \][/tex]
[tex]\[ x = \frac{12 + \sqrt{66}}{26} \][/tex]
To find the largest possible value of [tex]\( x \)[/tex], we select the larger of the two solutions:
[tex]\[ x = \frac{12 + \sqrt{66}}{26} \][/tex]
So, the largest possible value of [tex]\( x \)[/tex] is [tex]\( \frac{12 + \sqrt{66}}{26} \)[/tex].
