Maximum Box Volume problem solved with derivative and quadratic equation

Max Prism Volume Solution
Volume = ½ Base * Height * Length Length is 10. Divide triangle in two half to form two right triangles. A = ½ Base * Height* Substitute trigonometric expressions A(θ) = ½ * 4 sinθ * 4 cosθ A(θ) = 8 sinθ cosθ Use trigonometric identity (2 sinθ cosθ = sin2θ) A(θ) = 4 sin2θ Take derivative 0 = 8 cos2θ 0 = cos2θ cos2θ equals zero when 2θ = $\frac{π}{2}$ 2θ = $\frac{π}{2}$ θ = $\frac{π}{4}$ Substitute $\frac{π}{4}$ into area formula A(θ) = 4 sin2θ A( $\frac{π}{4}$ ) = 4 sin(2 * $\frac{π}{4}$ ) = 4 sin( $\frac{π}{2}$ ) Since sin( $\frac{π}{2}$ ) = 1: A = 4( $\frac{π}{2}$ ) = 4 * 1 = 4 This is are of the right triangle. Area of the isosceles triangle is double. A = 4 + 4 = 8 Volume = Area of Triangle * Length V = 8 * 10 = 80 cubic inches
177

Max Prism Volume
Solution

Volume = ½ Base * Height * Length
Length is 10. Divide triangle in two
half to form two right triangles.
Triangular end of prism.

A = ½ * Base * Height
Substitute trigonometric expressions
A(θ) = ½ * 4 sinθ * 4 cosθ
A(θ) = 8 sinθ cosθ

Use trigonometric identity
(2 sinθ cosθ = sin2θ)
A(θ) = 4 sin2θ

Take derivative
0 = 8 cos2θ
0 = cos2θ

cos2θ equals zero when 2θ = $\frac{π}{2}$
2θ = $\frac{π}{2}$
θ =

\frac{π}{4}

Substitute

\frac{π}{4}

into area formula
A(θ) = 4 sin2θ
A( $\frac{π}{4}$ ) = 4 sin(2 * $\frac{π}{4}$ ) = 4 sin( $\frac{π}{2}$ )
Since sin( $\frac{π}{2}$ ) = 1:
A = 4( $\frac{π}{2}$ ) = 4 * 1 = 4

This is are of the right triangle.
Area of the isosceles triangle is double.
A = 4 + 4 = 8

Volume = Area of Triangle * Length
V = 8 * 10 = 80 cubic inches

177

Max Prism VolumeSolution

Max Prism Volume
Solution