Application of Derivatives Part 2

Let the side of the square to be cut off be x‌cm.
Then, the length and breadth of the box will be (30−2x)‌cm each and the height of the box is x‌cm therefore,The volume V(x) of the box is given by
‌V(x)=x(30−2x)2
‌‌

dv

dx

=(30−2x)2+2x×(30−2x)(−2)
‌0=(30−2x)2−4x(30−2x)
‌0=(30−2x)[(30−2x)−4x]
‌0=(30−2x)(30−6x)
‌x=15,5
‌x≠15‌ (Not possible) ‌
‌{∴V=0}‌‌
(Not possible)
Surface area without top of the box =ℓb+2(bh+hℓ)
‌=(30−2x)(30−2x)+2[(30−2x)x+(30−2x)x]
‌=[(30−2x)2+4{(30−2x)x}.
‌=[(30−10)2+4(5)(30−10)]
‌=400+400
‌=800cm2