If the length, breadth and height of the box is denoted by a, b and h respectively, then V=a×b×h =32, and so h=32/ab. Now we have to maximize the surface area (lateral and the bottom) A = (2ah+2bh)+ab =2h(a+b)+ab = [64(a+b)/ab]+ab =64[(1/b)+(1/a)]+ab.
We treat A as a function of the variables and b and equating its partial derivatives with respect to a and b to 0. This gives {-64/(a^2)}+b=0, which means b=64/a^2. Since A(a,b) is symmetric in a and b, partial differentiation with respect to b gives a=64/b^2, ==>a=64[(a^2)/64}^2 =(a^4)/64. From this we get a=0 or a^3=64, which has the only real solution a=4. From the above relations or by symmetry, we get b=0 or b=4. For a=0 or b=0, the value of V is 0 and so are inadmissible. For a=4=b, we get h=32/ab =32/16 = 2.
Therefore the box has length and breadth as 4 ft each and a height of 2 ft.
Answer:
eight
Step-by-step explanation:
Answer:
Blue Box / First answer choice
Step-by-step explanation:
Plug in d1:8.6 and d2:10.2 into d1 x d2/2 to find the area.
d1 x d2/2
8.6 x 10.2/2
87.72/2
Area=43.86
Answer:
C
Step-by-step explanation:
5(4) over 6(4) makes more sense than the others do
The answer would be 36.75 units squared because the area of the big square is 196 units squared, then take away (minus) the area of the smaller square, 49 units squared is 147 units squared, then simply divide it by four because the grey area represents a fourth of the space in between the squares so that brings you to 36.75 units squared.
