The length script l, width w, and height h of a box change with time. At a certain instant the dimensions are script l = 7 m and

w = h = 5 m,
and script l and w are increasing at a rate of 6 m/s while h is decreasing at a rate of 4 m/s. At that instant find the rates at which the following quantities are changing.
(a) The volume.
m3/s

(b) The surface area.
m2/s

(c) The length of a diagonal. (Round your answer to two decimal places.)
m/s

I'll do the volume. You should be able to get the others. If not, come back and we can help you get unstuck.

v = lwh
dv/dt = wh dl/dt + lh dw/dt + lw dh/dt
plugging in the numbers,

dv/dt = 5m*5m*(6m/s) + 7m*5m*(6m/s) + 7m*5m*(-4m/s)
= 150+210-140=220 m^3/s

yes i already got parts a and b i do not know how to do part c sorry i should have specified that in my post

(a) The volume of a box can be calculated by multiplying its length, width, and height. Since the length (l) and width (w) are increasing at a rate of 6 m/s, we can use the product rule of derivatives to find the rate of change of volume:

dV/dt = (d/dt) (lwh)
= l(d/dt)(wh) + w(d/dt)(lh) + h(d/dt)(lw)
= l(w(dh/dt) + h(dw/dt)) + w(h(dl/dt) + l(dh/dt)) + h(l(dw/dt) + w(dl/dt))

Given that l = 7 m, w = h = 5 m, dw/dt = dl/dt = 6 m/s, and dh/dt = -4 m/s:

dV/dt = 7(5(-4) + 5(6)) + 5(5(6) + 7(-4)) + 5(7(6) + 5(6))
= 7(-20 + 30) + 5(30 - 28) + 5(42 + 30)
= 7(10) + 5(2) + 5(72)
= 70 + 10 + 360
= 440 m^3/s

Therefore, the volume is changing at a rate of 440 m^3/s.

(b) The surface area of a box can be calculated by summing the areas of its six faces. Since only the length (l) and width (w) are changing, we can find the rate of change of surface area using the product rule of derivatives:

dS/dt = (d/dt) (2lw + 2lh + 2wh)
= 2(d/dt)(lw) + 2(d/dt)(lh) + 2(d/dt)(wh)
= 2(w(dl/dt) + l(dw/dt)) + 2(h(dl/dt) + l(dh/dt)) + 2(w(dh/dt) + h(dw/dt))

Given that l = 7 m, w = h = 5 m, dl/dt = dw/dt = 6 m/s, and dh/dt = -4 m/s:

dS/dt = 2(5(6) + 7(6)) + 2(5(6) + 7(-4)) + 2(5(-4) + 5(6))
= 2(30 + 42) + 2(30 - 28) + 2(-20 + 30)
= 2(72) + 2(2) + 2(10)
= 144 + 4 + 20
= 168 m^2/s

Therefore, the surface area is changing at a rate of 168 m^2/s.

(c) The length of a diagonal can be calculated using the Pythagorean theorem:

diagonal^2 = length^2 + width^2 + height^2

Differentiating both sides of the equation with respect to time (t), we can find the rate of change of the length of the diagonal:

2(diagonal)(d(diagonal)/dt) = 2(length)(d(l)/dt) + 2(width)(d(w)/dt) + 2(height)(d(h)/dt)

Given that length (l) = 7 m, width (w) = height (h) = 5 m, and dl/dt = dw/dt = 6 m/s and dh/dt = -4 m/s:

2(diagonal)(d(diagonal)/dt) = 2(7)(6) + 2(5)(6) + 2(5)(-4)
(diagonal)(d(diagonal)/dt) = 84 + 60 - 40
(diagonal)(d(diagonal)/dt) = 104

Solving for d(diagonal)/dt:

d(diagonal)/dt = 104 / diagonal

Now, we need to find the value of the diagonal at that instant. Using the Pythagorean theorem:

diagonal^2 = 7^2 + 5^2 + 5^2
diagonal^2 = 49 + 25 + 25
diagonal^2 = 99
diagonal ≈ 9.95 m

Substituting the value of the diagonal into the equation for d(diagonal)/dt:

d(diagonal)/dt = 104 / 9.95
d(diagonal)/dt ≈ 10.45 m/s

Therefore, the length of the diagonal is changing at a rate of approximately 10.45 m/s.

To find the rates at which the quantities are changing, we can use the concepts of derivatives and related rates.

(a) The volume:
The volume of a box is given by the formula V = l * w * h. To find the rate at which the volume is changing, we can calculate the derivative of V with respect to time t.

V = l * w * h
Taking the derivative of both sides with respect to time:
dV/dt = (dl/dt) * w * h + l * (dw/dt) * h + l * w * (dh/dt)

Given the values:
dl/dt = 6 m/s (rate of change of length)
dw/dt = 6 m/s (rate of change of width)
dh/dt = -4 m/s (rate of change of height)

Substituting these values into the equation:
dV/dt = (6) * 5 * 5 + 7 * 6 * 5 + 7 * 5 * (-4)
dV/dt = 150 + 210 + (-140)
dV/dt = 220 m^3/s

Therefore, the rate at which the volume is changing is 220 m^3/s.

(b) The surface area:
The surface area of a box is given by the formula A = 2lw + 2lh + 2wh. To find the rate at which the surface area is changing, we can calculate the derivative of A with respect to time t.

A = 2lw + 2lh + 2wh
Taking the derivative of both sides with respect to time:
dA/dt = 2(dl/dt) * w * h + 2l * (dw/dt) * h + 2l * w * (dh/dt)

Given the values:
dl/dt = 6 m/s (rate of change of length)
dw/dt = 6 m/s (rate of change of width)
dh/dt = -4 m/s (rate of change of height)

Substituting these values into the equation:
dA/dt = 2(6) * 5 * 5 + 2 * 7 * 6 * 5 + 2 * 7 * 5 * (-4)
dA/dt = 600 + 420 + (-280)
dA/dt = 740 m^2/s

Therefore, the rate at which the surface area is changing is 740 m^2/s.

(c) The length of a diagonal:
The length of a diagonal of a box can be calculated using the Pythagorean theorem.
The diagonal d is given by the formula d = sqrt(l^2 + w^2 + h^2)

Taking the derivative of d with respect to time:
dd/dt = (1/2) * (l^2 + w^2 + h^2)^(-1/2) * (2l(dl/dt) + 2w(dw/dt) + 2h(dh/dt))

Given the values:
l = 7 m
w = 5 m
h = 5 m
dl/dt = 6 m/s (rate of change of length)
dw/dt = 6 m/s (rate of change of width)
dh/dt = -4 m/s (rate of change of height)

Substituting these values into the equation:
dd/dt = (1/2) * (7^2 + 5^2 + 5^2)^(-1/2) * (2 * 7 * 6 + 2 * 5 * 6 + 2 * 5 * (-4))
dd/dt = (1/2) * (99)^(-1/2) * (84 + 60 - 40)
dd/dt = (1/2) * (1/√99) * 104
dd/dt ≈ 0.933 m/s

Therefore, the rate at which the length of the diagonal is changing is approximately 0.933 m/s.

there are diagonals along each face, but thety probably want the diagonal of the brick:

d^2 = l^2 + w^2 + h^2
2d dd/dt = 2l dl/dt + 2w dw/dt + 2h hd/dt
when (l,w,h) = (7,5,5) d = sqrt(99) = 3sqrt(11)

3sqrt(11) dd/dt = 2*7*6 + 2*5*6 + 2*5*(-4)
3sqrt(11) dd/dt = 104
dd/dt = 104 / 3sqrt(11)