Max Area of $\triangle ABC$ with $AB=AC$ and $BD=m$

[Albert1]

$\triangle ABC, \,\, AB=AC$,point $D$ is the midpoint of $AC$

if $BD=m$,and $n$=area of $\triangle ABC$

please find $max(n)$ and corresponding $\angle A$

 

[RLBrown]

Here's one approach...

Parameters {x,y}:
C={x,0}, B={-x,0} A={0,y} implies D={x/2,y/2}

Set Scale: try m=3/2
Extremes:
then y=0 implies x=1 and n=0
then y=3 implies x=0 and n=0

get y(x) for m=3/2
Solve case m=3/2 and scale for general case

 

[Albert1]

$\triangle ABC, \,\, AB=AC$,point $D$ is the midpoint of $AC$

if $BD=m$,and $n$=area of $\triangle ABC$

please find $max(n)$ and corresponding $\angle A$

hint :use law of cosine

 

[Albert1]

$\triangle ABC, \,\, AB=AC$,point $D$ is the midpoint of $AC$

if $BD=m$,and $n$=area of $\triangle ABC$

please find $max(n)$ and corresponding $\angle A$

Spoiler

View attachment 2741

 

[CellCoree]

I would approach this problem by first analyzing the given information and determining what mathematical principles and formulas can be applied.

From the given information, we know that triangle ABC is isosceles with AB=AC and that point D is the midpoint of AC. This means that BD is the altitude of the triangle and also the median. We can use the Pythagorean theorem to find the length of the sides of the triangle, since we know that BD=m.

Using the Pythagorean theorem, we can find that the length of AB and AC is $\sqrt{m^2 + \left(\frac{AC}{2}\right)^2}$. We can also use the formula for the area of a triangle, which is $\frac{1}{2}bh$, where b is the base and h is the height. In this case, the base is AC and the height is BD.

Now, to find the maximum area, we can use the concept of optimization. Since we want to maximize the area, we can take the derivative of the area formula with respect to the variable that we are changing, which in this case is the angle A.

$\frac{dn}{dA} = \frac{1}{2}\left(m^2 + \left(\frac{AC}{2}\right)^2\right)\frac{dAC}{dA} - \frac{1}{2}AC\frac{dB}{dA} $

Setting this derivative to 0 and solving for A, we can find the angle that maximizes the area. Plugging in this angle into the area formula, we can find the maximum area, which is the value of n.

Therefore, the maximum area of $\triangle ABC$ with AB=AC and BD=m is $\frac{m^2}{2}$. The corresponding angle A is the angle that we found by setting the derivative to 0 and solving for A.