Maximum Box Volume problem solved with derivative and quadratic equation

Max Box Volume Solution
Volume = Length * Width * Height Let x be the size of corners cut out V(x) = x * (10 - 2x) * (10 - 2x) V(x) = 4x³ - 40x² + 100x Take the derivative. 0 = 12x² - 80x + 100 Apply quadratic equation. x = $\frac{80 \pm \sqrt{6400 - 4800}}{24}$ = $\frac{5}{3}$ and 5 Apply possible values into volume formula to find the largest volume. V(x) = x * (10 - 2x) * (10 - 2x) V(5) = 5 (10 - 10) (10 - 10) = 0 V( $\frac{5}{3}$ ) = $\frac{5}{3}$ (10 - $\frac{10}{3}$ ) (10 - $\frac{10}{3}$ ) = $\frac{2000}{27}$ Maximum dimensions: $\frac{5}{3}$ $\frac{20}{3}$ * $\frac{20}{3}$*
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Max Box Volume
Solution

Volume = Length * Width * Height
Let x be the size of corners cut out

Drawing of sheet of metal folded into box

V(x) = x * (10 - 2x) * (10 - 2x)
V(x) = 4x³ - 40x² + 100x

Take the derivative.
0 = 12x² - 80x + 100

Apply quadratic equation.
x = $\frac{80 \pm \sqrt{6400 - 4800}}{24}$ = $\frac{5}{3}$ and 5

Apply possible values into volume
formula to find the largest volume.
V(x) = x * (10 - 2x) * (10 - 2x)

V(5) = 5 (10 - 10) (10 - 10) = 0

V( $\frac{5}{3}$ ) = $\frac{5}{3}$ (10 - $\frac{10}{3}$ ) (10 - $\frac{10}{3}$ ) = $\frac{2000}{27}$

Maximum dimensions: $\frac{5}{3}$ * $\frac{20}{3}$ * $\frac{20}{3}$

193

Max Box VolumeSolution

Max Box Volume
Solution