Question

The dimensions of a box are 100 cm,80 cm, 60 cm. The area of four walls is _____ sq.cm

Answer 1

Answer:

SOLUTION: Let the dimensions of the box be l, w and h (for length, width and height).

The surface area is then:

S(l, w, h) = 2lw + 2wh + 2lh = 2(lw + wh + lh)

The change in area can be written as:

∆S ≈ dS = Sl dl + Sw dw + Sh dh

where the partial derivatives are evaluated at l = 80, w = 60 and h = 50, and

dl = dw = dh = 0.2.

The partial derivatives are computed:

Sl = 2(w + h) = 220 Sw = 2(l + h) = 260 Sh = 2(l + w) = 280

Substituting these in for dS,

dS = 220 · 0.2 + 260 · 0.2 + 280 · 0.2 = 152 cm2

35. Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm

and height 12 cm if the tin is 0.04 cm thick.

SOLUTION: The volume of the can is

V (r, h) = πr2h

Using differentials,

∆V ≈ dV = Vr dr + Vh dh

with r = 4 and h = 12, dr = 0.04 and dh = 2 · 0.04 = 0.08.

Compute the partial derivatives out and we get:

dV = 2πrh dr + πr2

dh

Substitute in the numerical values:

dV = 3.84π + 1.28π ≈ 16.08 cm3

(The book rounds off a little too much)

Step-by-step explanation:

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