Question
The perimeter of an isosceles triangle is 16 cm. Do the following:
(A) Make a drawing of the problem.
(B) What are the dimensions of the sides and height for the maximum area?
(C) What is the maximum area?
Answer/Explanation
(A)
Given:P=16cm
P=2x+base
16=2x+base
base=16-2x
\(\frac{base}{2}=8-x\)
(B)\(x^{2}=h^{2}+(8-x)^{2}\)
\(x^{2}=h^{2}+(64-16x+x^{2})\)
\(x^{2}=h^{2}+64-16x+x^{2}\)
\(0=h^{2}+64-16x\)
\(h^{2}=16x-64\)
\(h^{2}=16(x-4)\)
\(h=4\sqrt{(x-4)}\)
\(A=\frac{1}{2}bh\)
\(A=\frac{1}{2}(16-2x)(4\sqrt{x-4})\)
\(A=(8-x)[2(x-4)^{\frac{1}{2}}]\)
\(\frac{\mathrm{d} A}{\mathrm{d} x}=\frac{1}{2}(2)(x-4)\frac{1}{2}(8-x)+(2)(x-4)^{\frac{1}{2}}(-1)\)
\(\frac{\mathrm{d} A}{\mathrm{d} x}=(x-4)^{\frac{1}{2}}(8-x)-(2)(x-4)^{\frac{1}{2}}\)
\(\frac{\mathrm{d} A}{\mathrm{d} x}=\frac{(8-x)-(2)(x-4)}{\sqrt{x-4}}\)
\(\frac{\mathrm{d} A}{\mathrm{d} x}=\frac{8-x-2x+8}{\sqrt{x-4}}\)
\(\frac{\mathrm{d} A}{\mathrm{d} x}=\frac{16-3x}{\sqrt{x-4}}\)
\(0=\frac{16-3x}{\sqrt{x-4}}\)
x=4 leads toan undefined solution
0=16-3x
3x=16
\(x=\frac{16}{3}=5.33cm\)
\(base=(16-2x)=\left [ 16-(2)\left ( \frac{16}{3} \right ) \right ]=\left [ 16-\frac{32}{3} \right ]=\left [ \frac{48}{3}-\frac{32}{3} \right ]=\frac{16}{3}=5.33cm\)
Therefor,the triangle is an equilateral triangle.
\(h=4\sqrt{(x-4)}=4\sqrt{\left ( \frac{16}{3}-4 \right )}=4\sqrt{\left ( \frac{16}{3} -\frac{12}{3}\right )}=4\sqrt{\frac{4}{3}}=\frac{8\sqrt{3}}{3}=4.62 cm\)
(C)\(A= \frac{1}{2}bh\)
\(A=\frac{1}{2}\left ( \frac{16}{3} \right )\left ( 4\sqrt{\frac{4}{3}} \right )\)
\(A=\left [ (2)\left ( \frac{16}{3} \right ) \sqrt{\frac{4}{3}}\right ]\)
\(A=\left [ \frac{32}{3}\sqrt{\frac{4}{3}} \right ]\)
A=(10.67cm)(1.16cm)
\(A=12.3cm^{2}\)