Answer to Question #218951 in Calculus for jamie

Question

The surface area A of an average human body is established to be given by A(w,h)=16.54w^0.75h^0.25 where w is the weight and h is the height.

1. Find A_W(65,57) and A_h(65,57), interpret the results.
2. For a 67 pounds child whose height is 57 inches determine the change in area A if weight decreases by 5 pounds and height increases by3 inches.

Answer 1

$A(w,h)=16.54w^{0.75}h^{0.25}$

$A_w(w,h)=\frac{\partial A}{\partial w}= 16.54\cdot 0.75w^{-0.25}h^{0.25}=12.405(h/w)^{1/4}$

$A_w(65,57)=12.405(57/65)^{1/4}=12.0043$

This equality means that the surface area A of an average human body with weight and heght close to (65, 57) changes by approximately 12.0043 per each pound.

$A_h(w,h)=\frac{\partial A}{\partial h}= 16.54\cdot 0.25w^{0.75}h^{-0.75}=4.135(h/w)^{-3/4}$

$A_h(65,57)=4.135(57/65)^{-3/4}=4.653$

This equality means that the surface area A of an average human body with weight and heght close to (65, 57) changes by approximately 4.653 per each inch.

If $w=65$, $h=57$, $\Delta w=5$ and $\Delta h=3$, then the change in area A will be approximately equal to

$\Delta A\approx A_w\Delta w + A_h\Delta h = 12.0043\cdot 5+4.653\cdot3=72.0258$