### Math Notes

Subjects

#### Differential Calculus Solutions

##### Topics || Problems

What should be the length of the third side of an isosceles triangle, if the congruent sides is 10 and the area is at maximum?

Solution:

$$A = \frac{bh}{2}$$

$$10^2 = h^2 +(\frac{b}{2})^2$$

$$h = \sqrt{100- (\frac{b}{2})^2}$$

$$A = \frac{(b)\sqrt{100- (\frac{b}{2})^2}}{2}$$

$$A = \frac{\sqrt{100b^2- \frac{b^4}{4}}}{2}$$

$$\frac{2dA}{dx} = \frac{200b-b^3}{2\sqrt{100b^2-\frac{b^4}{4}}}$$

$$0 = 200b-b^3$$

$$b = 14.14$$ units

Thus the length of the third side is:

$$b = 14.14$$ units