SOLUTION:
Case: Area of triangle (maximum value) application
Given: expressions for the base and height of the triangle
length of base = x
height = 2 (24-2x)
Required: Equate Area, A(x) = (1/2) bh. Find the maximum area
Method:
The Area, A(x) is given as:
[tex]\begin{gathered} A(x)=\text{ }\frac{bh}{2} \\ A(x)=\text{ }\frac{x\times2(24-2x)}{2} \\ A(x)=\text{ }x(24-2x) \\ A(x)=\text{ }24x-2x^2 \\ \text{Differentiating;} \\ A^{\prime}(x)=\text{ }24-4x^{} \\ \text{The maximum area is obtained where A'(x)=0} \\ 0=\text{ }24-4x^{} \\ 4x=24 \\ \text{Divide both sides by 4} \\ x=\text{ 6} \end{gathered}[/tex]
The maximum area therefore is A(x)=
[tex]\begin{gathered} A(x)=\text{ }24x-2x^2 \\ \text{Where x=6} \\ A(6)=\text{ }24(6)-2(6)^2 \\ A(6)\text{ = 144 -2(36)} \\ =\text{ 14}4-72^{} \\ A=\text{ 72} \end{gathered}[/tex]
Final answer:
The maximum potential area is 72 square inches.Option (D)
