Answer :
a) 331776 [tex]cm^3[/tex] units are being optimized.
b) The dimensions would be if the length and width were equal to 373248 [tex]cm^3[/tex] units.
Using the Lagrange method,
a) Any suitcase should be less than or equivalent to 216 centimeters in length, breadth, and height.
L + W + H = 216
L = W if the width and length are equal.
Let's replace L with W to create this new equation.
2W + H = 216
Left-justify H by putting H = 216 - 2W.
Volume, V = L × W × H = [tex]W^2[/tex] × (216 - 2W) = 216[tex]W^2[/tex] - 2[tex]W^3[/tex]
Similar to your previous inquiry, you locate the first derivative & set it to 0 whenever you wish to maximize anything.
dV/dW = 432W - 6[tex]W^2[/tex]
432W - 6[tex]W^2[/tex] = 0
divide 6 on both sides,
72 w - [tex]w^2[/tex] = 0
W = 72
Reconnect it, and you'll see that W = L = H = 72 cm.
Therefore maximum volume = [tex]72^3[/tex] = 373248 [tex]cm^3[/tex] units
b) The only thing that changed is L = 2W.
Given L + W + H = 216
Now 3W + H = 216
V = L × W × H = 2W × W × (216 - 3W) = 432[tex]W^2[/tex] - 6[tex]W^3[/tex]
V prime = 864W - 18[tex]W^2[/tex]
W = 48 is obtained by setting V prime = 0 & solving it by reducing both sides by 18.
When you re-insert W, the values you receive are W = 48 cm, L = 96 cm, & H = 72 cm.
Therefore maximum volume = 48cm × 96cm × 72cm = 331776 [tex]cm^3[/tex] units.
Learn more about the Lagrange method at
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