Answer:
Step-by-step explanation:
Given that:
Let the others be = b
One of the bottom edge is five times longer than the others
i.e. L= 5b
The Volume of a box = L × b × h
V = 5b²h
where; V is constant
Differentiating with respect to t
[tex]2bh \bigg ( \dfrac{db}{dt} \bigg) + b^2\bigg(\dfrac{dh}{dt} \bigg) = 0[/tex]
2h(db) = -b(dh)
Therefore, the surface area = lb + 2bh + 2lh ; since there is no top.
A = 5b² + 2bh + 10bh
A = 5b² + 12bh
When A is minimum, [tex]\dfrac{dA}{dt}[/tex] = 0
10b(db) + 12b(dh) + 12h(db) = 0
[tex]10b(db) + 12b[\dfrac{-2h(db)}{b}] + 12h(db) = 0[/tex]
10b(db) + 12h(db) = 24h(db)
10b = 12h
5b = 6h
Recall that
l = 5b and L × b × h = V
Thus;
5b × b × [tex]\dfrac{5b}{6}[/tex] = V
[tex]b = 3\sqrt{\dfrac{6V}{25}}[/tex]
L = 5b
h = [tex]\dfrac{5b}{6}[/tex]
