Answer:
49
Step-by-step explanation:
∆ABC~ ∆XYZ
AB= 11 that is half of XY = 22
therefore XZ/2 = AC (base) = 32/2 = 16
and YZ /2 = BC(hypotenuse) = 44/2 = 22
(a= 11, b= 16, c= 22)
perimeter of a triangle
P = a+b+c
P= 11+16+22= 49
Lagrange multipliers:







(if

)

(if

)

(if

)
In the first octant, we assume

, so we can ignore the caveats above. Now,

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of

.
We also need to check the boundary of the region, i.e. the intersection of

with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force

, so the point we found is the only extremum.
Answer:
The fraction is 41/99.
Step-by-step explanation:
The steps are :






Answer: -8-10
Step-by-step explanation:
If the width is w and the length is l, then 2w-2=l and 2w+2l=72 (using the perimeter equation). Plugging 2w-2 in for l, we get 2w+(2w-2)*2=72 and 6w-4=72. Adding 4 to both sides, we get 6w=76. After that, we divide both sides by 6 to get 74/6=w. Since l=2w-2=136/6, we get (136/6)(74/6)=656.75=area