Answer to Question #214757 in Calculus for Maryam

Question

A closed rectangular box with a volume of 16 ft cube is to be made of three different materials. The cost of the material for the top and the bottom is $9 per sq. ft, the cost of the material for the front and the back is $8 per sq. ft. and the cost of the material for other two sides is $6 per sq. ft. Find the dimensions of the box (by two ways if possible) so that the cost of materials is minimized.

Answer 1

$\text{Let x = length, y = width and z = height. From the question wee have that}\\xyz=16-(1) \\\text{The total cost is given by }\\9(2xy)+8(2yz)+6(2xz)=c-(2)\\\text{From eqn(1), we have that $z=\frac{16}{xy}$}\\\text{We put the value of z in equation 2, therefore, we have}\\18xy+256x^{-1}+192y^{-1}=c\\\text{We differentiate c seperately with respect to x and y to obtain the following equations }\\18y-256x^{-2}=0-(3)\\18x-192y^{-2}=0-(4)\\\text{Using elimination method we have that}\\x=\frac{256}{192}y\\\text{Substituting the value of x in equation 3, we have}\\18(\frac{256}{192})y-192y^{-2}=0\\\text{Calculating y, we have y = 2,since x = $\frac{256}{192}y$ and z =$ \frac{16}{xy}$}\\\text{therefore x = 2.67, y = 2, z=3}$