Question

A measure of the adhesiveness of honey can be modeled as

f(x.y) = -125.48 + 4.26x + 4.85y -0.05x^2 - 0.14y^2

where x is the percentage of glucose and maltose and y is the percentage of moisture.

(a) calculate the absolute maximum of the adhesiveness of honey.

(b) the FDA restrictions for Grade A honey require that the combined percentages of glucose, maltose, and moisture equals 58%, what is the maximum measure of adhesiveness possible?

Answer 1

ANSWER:

(A). Given That f(x) = -125.48 + 4.26x + 4.85y - 0.05x^2 -0.14y^2

where 'x' is the percentage of glucose and maltose, and 'y' is the percentage of moisture.

now find critical number fx = 0 and fy = 0

fx = 4.26-0.1x = 0

=> x= 4.26 / 0.1 = 42.6

fy = 4.85-0.28y = 0

=> y = 4.85 / 0.28 = 17.32

fxx=-0.1 , fyy=-0.28 and fxy=fyx=0

now , fxx.fyy-(fxy)^2 = (-0.1)(-0.28)-0 = 0.028>0

so D>0, fxx<0 so f is maximum.

=> f(42.6,17.32) = -125.48+4.26*42.6+4.85*17.32-0.05(42.6)^2-0.14(17.32)^2

= -125.48+181.476+84.022-90.738-41.9975

= 7.28

so the absolute maximum of the adhesiveness of honey = 7.28

(B). g(x,y): x+y = 58

=> g(x,y)=x+y-58

4.26-0.1x = 4.85-0.28y => -0.1x = 4.85-0.28y-4.26

-0.1x = 0.59-0.28y => x = -5.9+2.8y

x+y = 58

=> -5.9+2.8y+y = 58

=> -5.9+3.8y = 58

=> 3.8y = 58+5.9 = 63.9

y = 63.9/3.8 = 16.8157

therefore, y= 16.8157

substitute y value in x+y=58

=> x = 58-y = 58-16.8157 = 41.1842

therefore, x = 41.1842

f(x,y)= f( 41.1842, 16.8157) = -125.48+4.26*41.1842+4.85*16.8157-0.05(41.1842)^2-0.14(16.8157)^2

= -125.48+175.44+81.55-84.8-39.58

= 7.13

therefore, maximum measure of adhesiveness of honey is 7.13