If 300 cm2 of material is available to make a box with a square base and an open top, find the maximum volume of the box in cubic centimeters. Answer to the nearest cubic centimeter without commas. For example, if the answer is 2,000 write 2000.
we have 300m^2 of material this box has a square base but variable height y so SA=x^2+4y. thus 300=x^2+4y we know V=x^2y and by fdt we know a critcal point exists when when V'=0 we need y, 300=x^2+4y => 300-x^2=4y and y=300/4-x^2/4 so V=x^2(75-x^2/4)=75x^2-1/4x^4 V'=150x-x^3=x(150-x^2)=0 if x=0 or x=sqrt150 Test these values, V=x^2(75-x^2/4) Clearly, [V=5625] and 300=x^2+4y=150+4(75-150/4)
