In the ma-and-pa grocery store, shelf-space is limited and must be used effectively to increase profit. two-cereal items, grano and wheatie, compete for a total shelf space of 60 ft2 . a box of grano occupies .2ft2 and a box of wheatie needs .4ft2 . the maximum daily demands of grano and wheatie are 200 and 120 boxes, respectively. a box of grano nets in $1.00 in profit and a box of wheatie $1.35. ma-and-pa thinks that because the unit profit of wheatie is 35% higher than that of grano, wheatie should be allocated 35% more space than grano, which amounts to allocating about 57% to wheatie and 43% to grano. what do you think?

No. 2/3 of the space to Grano and 1/3 of the space to Wheatie.

Step-by-step explanation:

Allocating about 57% to Wheatie and 43% to Grano, means to allocate 60(.57)=34.2 ft2 of Wheatie and 60(.43)=25.8 ft2 of Grano. In that case there would be 34.2/.4=85.5≈85 boxes of Wheatie and 25.8/.2=129 boxes of Grano. The total profit would be 129(1)+85(1.35)=$243.75

Best option:

200 Granos boxes and 50 Wheaties boxes on the shelf.

200(.2)=40ft^2 will allocate Granos boxes

50(.4)=20ft^2 will allocate Wheaties boxes.

This means that 40/60=2/3=66.6% of the space will allocate Granos boxes and 20/60=1/3=33.3% of the space will allocate Wheaties boxes.

The total profit would be 200(1)+50(1.35)=$267.5

EXTRA:

This is a optimization problem.

Let X1 be the number of Granos boxes

Let X2 be the number of Wheaties boxes

Objective:

Max Z=1(X1)+1.35(X2)

Subjecto to

0.2(X1)+0.4(X2)<=60,

X1<=200,

X2<=120,

X1,X2>=0.

You can solve it using the simplex method. Check the image for more details.

so the possibilities of the measurement of the side of the triangle are either it is longest side or not the longest side

when it is the longest side

s < 25 + 35s < 60

if it is not the longest side

35 < 25 + s10 < sso 10 < s < 60so the perimeter70 < p < 130