Let the number of large bookcases be x and number of small bookcases be y, then
Maximise P = 80x + 50y;
subkect to:
6x + 2y ≤ 24
x, y ≥ 2
The corner points are (2, 2), (2, 6), (3.333, 2)
For (2, 2): P = 80(2) + 50(2) = 160 + 100 = 260
For (2, 6): P = 80(2) + 50(6) = 160 + 300 = 460
For (3.333, 2): P = 80(3.333) + 50(2) = 266.67 + 100 = 366.67
Therefore, for maximum profit, he should produce 2 large bookcases and 6 small bookcases.
Recall that to get the x-intercepts, we set the f(x) = y = 0, thus
![\bf \stackrel{f(x)}{0}=-4cos\left(x-\frac{\pi }{2} \right)\implies 0=cos\left(x-\frac{\pi }{2} \right) \\\\\\ cos^{-1}(0)=cos^{-1}\left[ cos\left(x-\frac{\pi }{2} \right) \right]\implies cos^{-1}(0)=x-\cfrac{\pi }{2} \\\\\\ x-\cfrac{\pi }{2}= \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bf%28x%29%7D%7B0%7D%3D-4cos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%5Cimplies%200%3Dcos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%280%29%3Dcos%5E%7B-1%7D%5Cleft%5B%20cos%5Cleft%28x-%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%20%5Cright%29%20%5Cright%5D%5Cimplies%20cos%5E%7B-1%7D%280%29%3Dx-%5Ccfrac%7B%5Cpi%20%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0Ax-%5Ccfrac%7B%5Cpi%20%7D%7B2%7D%3D%0A%5Cbegin%7Bcases%7D%0A%5Cfrac%7B%5Cpi%20%7D%7B2%7D%5C%5C%5C%5C%0A%5Cfrac%7B3%5Cpi%20%7D%7B2%7D%0A%5Cend%7Bcases%7D)
Answer:
9/5
Step-by-step explanation:
solve the original equation for x
-4× + 3 = 1x +(-6)
-1x
-5x + 3 = -6
-3
-5x = -9
/-5
x = 9/5
CHECK
-4(9/5) + 3 = 1(9/5) +(-6)
-21/5 = -21/5
I'd do 9x4=36 because the 9 is from your money and the 4 is from your books. Now since you want to find out how much 4 books cost you just multiply that by 9
