The volume of a cube is increasing at a rate of 10 cm^3/min. How fast is the surface area increasing when the length of an edge 90 cm?

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Answer 1 · 2016-12-19T06:05:22

Surface area of the cube is increasing at a rate of #4/9# #(cm^2)/min#

If the length of an edge of a cube is #l# #cm.#,

its volume #V# is #l^3# and surface area #A# is #6l^2#.

Differentiating #V=l^3# w.r.t. time, we get

#(dV)/(dt)=3l^2(dl)/(dt)#

As #(dV)/(dt)=10# #(cm^3)/min#, wen #l=90#

#(dl)/(dt)=10/(3xx90^2)=1/2430#

As #A=6l^2#

#(dA)/(dt)=12lxx(dl)/(dt)=12xx90xx1/2430=4/9# #(cm^2)/min#