As the zoom lens in a camera moves in & out, the size of the rectangular image changes. If the current image is 8cm x 6cm, find the rate at which the area of the img is changing (dA/df) if the length of the image is changing at 0.5cm/s and width changing at 0.1cm/s

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As the zoom lens in a camera moves in & out, the size of the rectangular image changes. If the current image is 8cm x 6cm, find the rate at which the area of the img is changing (dA/df) if the length of the image is changing at 0.5cm/s and width changing at 0.1cm/s

Mathematics
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Area = LW ; dL/dt = .5; dW/dt = .1
dA/dt = dA/dL dL/dt or dA/dt = dA/dW dW/dt
dA/dt = L dW/dt + W dL/dt might be better :)

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Other answers:

dA/dt = L dW/dt + W dL/dt is 8 the length or the width? dA/dt = L dW/dt + W dL/dt
if length we do this: dA/dt = 8* .1 + 6 * .5 A' = .8 + 3.0 = 3.8 cm^2/sec
thanks alot

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