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. Use Pythagorean Theorem Solve for y x² + y² = 4² x² + y² = 16 y = $\sqrt{16 - x²}$ A = ½ * Base * Height Substitute expression for x and y A(x) = ½ * x * $\sqrt{16 - x²}$ Square formula to simplify A²(x) = ¼ x² (16 - x²) A²(x) = ¼ (16x² - x⁴) Take derivative 0 = ¼ (32x - 4x³) 0 = 8x - x³ 0 = x(8 - x²) 0 = 8 - x² x² = 8 x = $\sqrt{8}$ = 2 $\sqrt{2}$ y = $\sqrt{16 - x²}$ y = $\sqrt{16 - 8}$ = $\sqrt{8}$ = 2 $\sqrt{2}$ Height of isosceles triangle is 2 $\sqrt{2}$ Base is 2 $\sqrt{2}$ + 2 $\sqrt{2}$ = 4 $\sqrt{2}$ Length of prism = 10 Volume of prism Volume = ½ Base * Height * Length V = ½ * 4 $\sqrt{2}$ * 2 $\sqrt{2}$ * 10 V = ½ * 4 * 2 * 2 * 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.

Use Pythagorean Theorem
Solve for y
x² + y² = 4²
x² + y² = 16
y = $\sqrt{16 - x²}$

A = ½ * Base * Height
Substitute expression for x and y
A(x) = ½ * x * $\sqrt{16 - x²}$

Square formula to simplify
A²(x) = ¼ x² (16 - x²)
A²(x) = ¼ (16x² - x⁴)

Take derivative
0 = ¼ (32x - 4x³)
0 = 8x - x³
0 = x(8 - x²)
0 = 8 - x²
x² = 8
x = $\sqrt{8}$ = 2 $\sqrt{2}$

y = $\sqrt{16 - x²}$
y = $\sqrt{16 - 8}$ = $\sqrt{8}$ = 2 $\sqrt{2}$

Height of isosceles triangle is 2

\sqrt{2}

Base is 2

\sqrt{2}

+ 2

\sqrt{2}

= 4

\sqrt{2}

Length of prism = 10

Volume of prism
Volume = ½ Base * Height * Length
V = ½ * 4

\sqrt{2}

* 2

\sqrt{2}

* 10
V = ½ * 4 * 2 * 2 * 10 = 80 cubic inches

177

Max Prism VolumeSolution

Max Prism Volume
Solution