Volume, V = length x breadth x height
= (4x)(x)(h)
= 4x²h

h = V/4x²

Surface area, A
(open cuboid so we don't include the top surface)

= Rectangular base area + area of side walls

= length x breadth + 2 x length x height + 2 x breadth x height

= 4x(x) + 2(4x)(h) + 2(x)(h)
= 4x² + 8x(V/4x²) + 2x(V/4x²)
= 4x² + 4V/2x + V/2x
= 4x² + 5V/2x

(Shown)

Find x first when t = 4.

Method① Since x increases at a constant rate of 2cms-¹

dx/dt = 2

Integrate both sides with respect to t,

x = 2t + c , c is a constant

When x = 1, t = 0,

1 = 2(0) + c
c = 1

So x = 2t + 1

When t = 4, x = 2(4) + 1 = 9


Method ②

Given : the increase in x is constant at 2cm s-¹,

From t = 0 to t = 4,

4 seconds have elapsed. So increase in x = 2cm s-¹ × 4s
= 8cm

Since x = 1 when t = 0,
when t = 4, x = 1 + 8 = 9


When x = 9, and knowing that V is to be taken as 20 and h = V/4x²

h = 20/(4(9²))

h = 20/324

h = 5/81

Since V is taken to be 20,

A = 4x² + 5(20)/2x
A = 4x² + 50/x
A = 4x² + 50x-¹


dA/dx = 2(4x) + (-1)50x-²

= 8x - 50/x²


When x = 1.5,

dA/dx = 8(1.5) - 50/(1.5)²

= -10 2/9


We need to find dA/dt

dA/dt = dA/dx × dx/dt

= -10 2/9 × 2
= -20 4/9

So the total surface area is decreasing at 20 4/9 cm² per second when x = 1.5

Thank you!!!