Amanufacturerforaproductdeterminesthat,foritsfirst600unitsoftheproducts sold,theirprofitisGHS40perunit.Theprofitoneachoftheunitsbeyond600is decreasedby5pesewastimesthenumberofadditionalproductsproduced.Findthe profitmaximisingoutput.Notethatorderscannotbelessthan600units

To find the profit-maximising output, we need to first understand the profit function.

1. For the first 600 units, the profit is GHS 40 per unit. So, the total profit for the first 600 units is 600 * 40 = GHS 24,000.

2. For units beyond 600, the profit decreases by 5 pesewas times the number of additional products produced. Let's denote the number of additional products as x. So, the profit for each additional unit is 40 - 0.05x.

3. The total profit function P(x) is then the sum of the profit from the first 600 units and the profit from the additional units. That is, P(x) = 24,000 + (40 - 0.05x) * x.

4. To find the profit-maximising output, we need to find the value of x that maximises P(x). This is done by taking the derivative of P(x) with respect to x, setting it equal to zero, and solving for x.

5. The derivative of P(x) is P'(x) = 40 - 0.1x.

6. Setting P'(x) = 0 gives 40 - 0.1x = 0, which solves to x = 400.

7. Therefore, the profit-maximising output is 600 (the initial units) + 400 (the additional units) = 1,000 units.

Please note that this is a simplified model and actual business decisions should take into account other factors such as market demand, production capacity, and cost variations.

To find the profit-maximising output, we need to first understand the profit function.

1. For the first 600 units, the profit is GHS 40 per unit. So, the total profit for the first 600 units is 600 * 40 = GHS 24,000.

2. For the units beyond 600, the profit decreases by 5 pesewas times the number of additional products produced. Let's denote the number of additional products as x. So, the profit for each additional unit is 40 - 0.05x.

3. The total profit function P(x) is then the sum of the profit from the first 600 units and the profit from the additional units. That is, P(x) = 24,000 + (40 - 0.05x) * x.

4. To find the profit-maximising output, we need to find the maximum of this profit function. This occurs where the derivative of the profit function is zero.

5. The derivative of P(x) is P'(x) = 40 - 0.1x.

6. Setting P'(x) = 0 gives x = 400.

7. However, since orders cannot be less than 600 units, the profit-maximising output is 600 + 400 = 1,000 units.

So, the manufacturer should produce 1,000 units to maximise profit.