Answer to Question #243701 in Calculus for ZXCVC

Question

TOPIC: General Application of Derivatives. Draw the necessary figure and indicate the dimension given.

1. The perimeter of an isosceles triangle is 28 cm. What is the maximum area possible?

Answer 1

Let $x=$ base of an isosceles triangle, $y=$ its leg. Then the perimeter is

$$P=x+2y=28=>x=28-2y$$

By the Heron's formula the area $A$ of the triangle is

$$A=\sqrt{\dfrac{P}{2}(\dfrac{P}{2}-x)(\dfrac{P}{2}-y)(\dfrac{P}{2}-y)}$$

Substitute

$$A^2=F(y)=14(14-(28-2y))(14-y)^2$$

$$F(y)=28(y-7)(y-14)^2, 7<y<14$$

Find the critical number(s)

$$F'(y)=(28(y-7)(y-14)^2)'$$

$$=28(y-14)^2+28(2)(y-7)(y-14)$$

$$=28(y-14)(y-14+2y-14)$$

$$=28(y-14)(3y-28)$$

$$F'(y)=0=>28(y-14)(3y-28)=0$$

$$y_1=14, y_2=\dfrac{28}{3}$$

Critical numbers: $\dfrac{28}{3}, 14.$

$F(7)=0, F(14)=0$

$$F(\dfrac{28}{3})=28(\dfrac{28}{3}-7)(\dfrac{28}{3}-14)^2=\dfrac{38416}{27}$$

The function $F(y)$ has the absolute maximum with value of $\dfrac{38416}{27}$ on $[7, 14]$ at $y=\dfrac{28}{3}.$

$$x=28-2(\dfrac{28}{3})=\dfrac{28}{3}.$$

Therefore the absolute area with given perimeter has equiliteral triangle

$$x=y=y=\dfrac{28}{3}\ cm$$

$$Area=A=\dfrac{196\sqrt{3}}{9} \ cm^2$$