The area is tight, right?

$8, 15, 17$ makes a right angle triangle, which has the maximum possible altitude of $8$ given its base of $15$ .

Since $8^2 + 15^2 = 17^2$ , by the converse of Pythagorean's Theorem, the first triangle is a right triangle where the angle between the $8$ and $15$ sticks is a right angle, so its area is $A_1 = \frac{1}{2} \cdot 8 \cdot 15 = 60$ .

Replacing the $17$ stick with a different length would cause the angle $x$ between the $8$ and $15$ sticks to no longer be a right angle, and so its area is $A_2 = \frac{1}{2} \cdot 8 \cdot 15 \sin x = 60 \sin x$ . Since $0 < x < 180°$ and $x \neq 90°$ , $\sin x < 1$ , and so $60 \sin x < 60$ , or $A_2 < A_1$ .

Therefore, you cannot replace the stick of length $17$ with one of a different length in order to make the area greater.

Area of a triangle is given by $A=\dfrac {ab\sin \theta}2$ , where $\theta$ is the measure of angle between sides of lengths $a$ and $b$ . Fixing $a=15$ , $b=8$ and $\theta$ between the two sides. Then $A$ is maximum when $\sin \theta = 1$ , that is $\theta = 90^\circ$ . We note that $\theta = 90^\circ$ when the third side $c=17$ .

Therefore, no , we cannot replace the stick of length 17 with one of a different length to make the area greater.

Since $17^2=8^2+15^2$ , $\theta=90^\circ$ .
Replacing the stick with length $17$ only changes $\theta$ , but the area equals $\frac{8\cdot15\sin\theta}2$ , which is maximum at $\theta=90^\circ$ .

Solving as optimization problem utilizing Heron's formula

$F = \sqrt{s (s-a) (s-b) (s-c)}$ with $s = \frac{a+b+c}{2}$

Lets state that a is the variable side length and b and c are constant, $b=15$ and $c=8$ .

$F = \frac{1}{4} \sqrt{(a+ (b+c)) (-a+ (b+c))(a+ (b-c))(a- (b-c))} = \frac{1}{4} \sqrt{-(a+ d) (a-d)(a+ e)(a- e)}$ with $d=b+c$ and $e=b-c$

$F= \frac{1}{4} \sqrt{-(a^2 - d^2) (a^2 - e^2)} = \frac{1}{4} \sqrt{-a^4 + a^2 (d^2+e^2) - d^2 e^2)} = \frac{1}{4} \sqrt{-x^2 + x (d^2+e^2) - d^2 e^2)}$ with $x=a^2$

$\frac{\partial F}{\partial x} = \frac{1}{8} \frac{-2 x + d^2 e^2}{F}$

$\frac{\partial F}{\partial x} = 0$ for maximum area $F \rightarrow x = \frac{d^2+e^2}{2} \rightarrow a = \sqrt{\frac{d^2+e^2}{2} }$ (only positive real solutions of interest).

$d=23, e = 7 \rightarrow a = \sqrt{\frac{23^2+7^2}{2} } = \sqrt{289} = 17$ , therefore no greater area possible

The area of any triangle can be calculated through $\frac{1}{2}$ $ab$ $sinC$ . As the sides $a$ and $b$ are set (as $15$ and $8$ ) in order to increase the area, we must increase the value of $sinC$ .

(In the following calculations, side $a$ represents the 15 unit side, side $b$ represents the 8 unit side, and side $c$ represents the 17 unit side. Angle $C$ is the angle found directly opposite side $c$ )

To calculate what sin $C$ is, we must first calculate what the angle $C$ is. We can use the law of cosines to do this: $cosC=$ $\frac{a^2+b^2 - c^2}{2ab}$ Substituting in the variables, this gives us $cosC=$ $\frac{15^2+8^2 - 17^2}{2X15X8}$ which gives a result of $0$

To calculate the angle $C$ from $cosC$ we must use inverse $cos$ : $cos^-0=90$ therefore $C = 90$

Since $sin90 = 1$ and $1$ is the maximum value of the function of $sin$ , we can therefore conclude that no changes can be made to side $c$ to increase the area of the triangle.

Nope. If you increase the length of one side you have to change the length of at least one other side.

The area of the triangle is half of the area of the parallelogram. Regardless of what the third side is, a parallelogram can be made by drawing lines parallel to those of length 8 and 15, and the area of the triangle would always be 60.