A croissant shop produces two products: bear claws (B) and almond filled croissants (C). Each bear claw requires 6 oz of flour, 1 oz of yeast, and 2 TS of almond paste. An almond filled croissant requires 3 oz of flour, 1 oz of yeast, and 4 TS of almond paste. The company has 6,600 oz of flour, 1,400 oz of yeast, and 4,800 TS of almond paste available for today's production run.

Bear claw profits are 20 cents each, and almond filled croissants profits are 30 cents each.

How many units of each should be produced to maximize profits?

What id the maximized profit?

A. B = 400; C = 1000; Max Z = $380

B. B = 500; C = 900; Max Z = $350

C. B = 250; C = 1500; Max Z = 425

D. B = 300; C = 750; Min Z = $380

letter A: B = 400; C = 1000; Max Z = $380

Step-by-step explanation:

bear claws (B) need 6 oz of flour (f), 1 oz of yeast (y), 2 ts of paste (p)

croissants (C) need 3 oz of flour (f), 1 oz of yeast (y), 4 ts of paste (p)

putting that information in equations, we have:

B = 6f + y + 2p

C = 3f + y + 4p

The total number of resources (R) are:

R = 6600f + 1400y + 4800p

Let's call M our total profit, "k1" the number of B produced and "k2" the number of C produced.

So, we can state that:

M = k1*0.2 + k2*0.3

The number of resources R2 will demand is calculated like this:

R2 = k1*B + k2*C = (6k1+3k2) f + (k1+k2) y + (2k1+4k2) p

using R2 and R, we can make some inequations:

6k1+3k2 2k1+k2 < = 2200

k1+k2 < = 1400

2k1+4k2 k1+2k2 < = 2400

if we try to maximize k2 (as it worths more), we will have k1 = 0 and k2 = 1200 (limited by p), but looking at the resources R, we will still have resources to use (f and y). Looking at B and C expressions, we see that removing one C gives enough 'p' to make 2 B, which is a good trade (as 2B worths 0.4 cents, and 1C worths 0.3 cents). we have 200 'y' remaining, so doing this 200 times give us k1 = 400 and k2 = 1000, and the only resource remaining will be some of 'f'.

calculating the profit M, we have:

M = 400*0.2 + 1000*0.3 = 380$

the right answer is letter A.