Question

Find the area of the largest rectangle that can be inscribed in a right triangle with legs of lengths 4 cm and 5 cm if two sides of the rectangle lie along the legs.

Answer 1

Answer:

$A=3 cm^{2}$

Step-by-step explanation:

We need to maximize the area of the rectangle inside the triangle, in this case, it is:

$A=(4-x)y$

the length of this rectangle is (4-x) and width is y.

No, we need to represent y in terms of x. We can use this ratio property here:

$\frac{4}{3}=\frac{x}{y}$

Let's solve it for y:

$y=\frac{3x}{4}$

Then the area will be:

$A=(4-x)\frac{3x}{4}$

$A=3x-\frac{3x^{2}}{4}$

Now, let's take the derivative of A whit respect to x.

$\frac{dA}{dx}=3-\frac{3x}{2}$

And equal to zero to get the maximum value of x and solve it for x.

$0=3-\frac{3x}{2}$

$x=2$

Therefore, the area of largest rectangle will be

$A=3(2)-\frac{3(2)^{2}}{4}$

$A=3 cm^{2}$

I hope it helps you!